Exercise 1: Choosing u and dv: the new integral is the judge

Integration by parts is the product rule read backwards: from $(uv)' = u'v + uv'$ comes $\int u\,dv = uv - \int v\,du$ . The formula does not compute anything. It TRADES the integral you have for another one, $\int v\,du$ , and the whole skill is to choose $u$ so that the new integral is easier than the old one.

Two requirements decide the choice: $dv$ must be something you can integrate, and $u$ should be the factor that becomes SIMPLER when differentiated. In every part, write the four pieces $u$ , $du$ , $dv$ , $v$ before the formula.

a) Compute $\int x e^{2x}\,dx$ with $u = x$ and $dv = e^{2x}\,dx$ .
b) Redo the first step with the opposite choice, $u = e^{2x}$ and $dv = x\,dx$ . Is the resulting equation false? Why is it useless?
c) Compute $\int x^3 \ln x\,dx$ . Explain why here the polynomial must be $dv$ and not $u$ .
d) Compute $\int x\, 2^x\,dx$ , and name the two wrong antiderivatives of $2^x$ that students write for $v$ .
e) Check the answers to a) and c) by differentiating them.

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Answers

a) $\int x e^{2x}\,dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C = \frac{e^{2x}(2x - 1)}{4} + C$
b) $\frac{x^2 e^{2x}}{2} - \int x^2 e^{2x}\,dx$ : true, but the power went UP.
c) $\int x^3 \ln x\,dx = \frac{x^4 \ln x}{4} - \frac{x^4}{16} + C$
d) $\int x\,2^x\,dx = \frac{x\,2^x}{\ln 2} - \frac{2^x}{(\ln 2)^2} + C$ ; wrong $v$ : $2^x$ and $\frac{2^{x+1}}{x+1}$ .
e) Both derivatives give back the integrand: $x e^{2x}$ and $x^3 \ln x$ .

a) $u = x$ , $du = dx$ ; $dv = e^{2x}\,dx$ , $v = \frac{1}{2}e^{2x}$ (a substitution you do at sight: the derivative of $\frac{1}{2}e^{2x}$ is $e^{2x}$ ). Then $\int x e^{2x}\,dx = \frac{x e^{2x}}{2} - \int \frac{1}{2}e^{2x}\,dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C = \frac{e^{2x}(2x - 1)}{4} + C$ . The trade was good: the new integral, $\int \frac{1}{2}e^{2x}\,dx$ , has lost the factor $x$ and is immediate. That is the test of a choice, and it can be run BEFORE finishing the computation: differentiating $u = x$ gives $1$ , and the power of $x$ disappears.

b) $u = e^{2x}$ , $du = 2e^{2x}\,dx$ ; $dv = x\,dx$ , $v = \frac{x^2}{2}$ . The formula gives $\int x e^{2x}\,dx = \frac{x^2 e^{2x}}{2} - \int \frac{x^2}{2} \cdot 2e^{2x}\,dx = \frac{x^2 e^{2x}}{2} - \int x^2 e^{2x}\,dx$ . This equation is TRUE: parts never produces a false identity. It is useless because the new integral carries $x^2$ where the old one carried $x$ : the trade went the wrong way, and applying parts again with the same choice would give $x^3$ , then $x^4$ , forever. When the integral you get is worse than the one you had, stop after one line and swap $u$ and $dv$ : that decision, written down, is worth marks on its own.

c) $u = \ln x$ , $du = \frac{dx}{x}$ ; $dv = x^3\,dx$ , $v = \frac{x^4}{4}$ . Then $\int x^3 \ln x\,dx = \frac{x^4 \ln x}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x}\,dx = \frac{x^4 \ln x}{4} - \int \frac{x^3}{4}\,dx = \frac{x^4 \ln x}{4} - \frac{x^4}{16} + C$ . Here the polynomial must be $dv$ , against the habit built in part a). Taking $dv = \ln x\,dx$ would require an antiderivative of $\ln x$ just to START, and would leave $\int 3x^2(x \ln x - x)\,dx$ , which contains $x^3 \ln x$ again. Differentiating $\ln x$ , on the other hand, turns it into $\frac{1}{x}$ , which cancels against $x^4$ : the logarithm is the factor that simplifies, so it is $u$ . The rule of thumb LIATE (logarithm, inverse trigonometric, algebraic, trigonometric, exponential, in that order of preference for $u$ ) says the same thing, but the reason is the new integral, not the mnemonic.

d) $u = x$ , $du = dx$ ; $dv = 2^x\,dx$ , $v = \frac{2^x}{\ln 2}$ , because $\frac{d}{dx}2^x = 2^x \ln 2$ . Then $\int x\,2^x\,dx = \frac{x\,2^x}{\ln 2} - \frac{1}{\ln 2}\int 2^x\,dx = \frac{x\,2^x}{\ln 2} - \frac{2^x}{(\ln 2)^2} + C$ . The two classic wrong values of $v$ : $v = 2^x$ , copied from $e^x$ (its derivative is $2^x \ln 2$ , off by the factor $\ln 2 \approx 0.69$ ), and $v = \frac{2^{x+1}}{x + 1}$ , the power rule applied to a variable EXPONENT, which is not a power function at all. Both errors propagate: $v$ appears twice, once in $uv$ and once in $\int v\,du$ , which is why the correct answer carries $\ln 2$ to the first power AND to the second.

e) For a): $\frac{d}{dx}\left[\frac{e^{2x}(2x - 1)}{4}\right] = \frac{2e^{2x}(2x - 1) + 2e^{2x}}{4} = \frac{4x e^{2x}}{4} = x e^{2x}$ . For c): $\frac{d}{dx}\left[\frac{x^4 \ln x}{4} - \frac{x^4}{16}\right] = x^3 \ln x + \frac{x^4}{4} \cdot \frac{1}{x} - \frac{x^3}{4} = x^3 \ln x$ . Both are confirmed. This check costs thirty seconds and is the only way, on an exam without a calculator, to know that an antiderivative is right; it catches a lost factor $\frac{1}{2}$ or a sign in $-\int v\,du$ immediately. Notice that the product rule, used to check, is exactly the rule that parts runs backwards.

Exercise 2: One factor only: ln x, arctan x and arcsin x with dv = dx

Some integrals have a single factor, and it is a function whose DERIVATIVE you know but whose antiderivative is not in the table: $\ln x$ , $\arctan x$ , $\arcsin x$ . Parts still applies, with the invisible partner $dv = dx$ and $v = x$ : the trade replaces the function by its derivative, which is algebraic.

The figure shows the graph of $y = \ln x$ and the region $A$ under it between $x = 1$ and $x = e$ .

a) Compute $\int \ln x\,dx$ .
b) Compute $\int \arctan x\,dx$ , naming the substitution used for the new integral.
c) Compute $\int_0^{1/2} \arcsin x\,dx$ exactly.
d) Compute the area of the region $A$ on the figure, and check that the value is plausible by comparing $A$ with a rectangle and a triangle that you can read on the figure.
e) Three students give $\int \ln x\,dx = \frac{1}{x} + C$ , $\frac{(\ln x)^2}{2} + C$ and $x \ln x + C$ . Refute each one by differentiating, and name the error behind it.

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Answers

a) $\int \ln x\,dx = x \ln x - x + C$
b) $\int \arctan x\,dx = x \arctan x - \frac{1}{2}\ln(1 + x^2) + C$
c) $\int_0^{1/2} \arcsin x\,dx = \frac{\pi}{12} + \frac{\sqrt 3}{2} - 1$
d) $A = 1$ , between the triangle $\frac{e - 1}{2}$ and the rectangle $e - 1$ .
e) Derivatives: $-\frac{1}{x^2}$ , $\frac{\ln x}{x}$ , $\ln x + 1$ ; none is $\ln x$ .

a) $u = \ln x$ , $du = \frac{dx}{x}$ ; $dv = dx$ , $v = x$ . Then $\int \ln x\,dx = x \ln x - \int x \cdot \frac{1}{x}\,dx = x \ln x - \int 1\,dx = x \ln x - x + C$ . The trade replaced the logarithm, which we cannot integrate directly, by its derivative $\frac{1}{x}$ , and the $v = x$ swallowed it. Check: $\frac{d}{dx}(x \ln x - x) = \ln x + 1 - 1 = \ln x$ .

b) $u = \arctan x$ , $du = \frac{dx}{1 + x^2}$ ; $dv = dx$ , $v = x$ . Then $\int \arctan x\,dx = x \arctan x - \int \frac{x}{1 + x^2}\,dx$ . The new integral is a substitution: $w = 1 + x^2$ , $dw = 2x\,dx$ , so $\int \frac{x}{1 + x^2}\,dx = \frac{1}{2}\int \frac{dw}{w} = \frac{1}{2}\ln(1 + x^2)$ , with no absolute value since $1 + x^2 > 0$ . Final answer: $x \arctan x - \frac{1}{2}\ln(1 + x^2) + C$ . Check: $\arctan x + \frac{x}{1 + x^2} - \frac{1}{2} \cdot \frac{2x}{1 + x^2} = \arctan x$ . The substitution is only a TOOL here; the structural move is $dv = dx$ .

c) $u = \arcsin x$ , $du = \frac{dx}{\sqrt{1 - x^2}}$ ; $dv = dx$ , $v = x$ . Then $\int_0^{1/2} \arcsin x\,dx = \left[x \arcsin x\right]_0^{1/2} - \int_0^{1/2} \frac{x}{\sqrt{1 - x^2}}\,dx$ . The bracket is $\frac{1}{2} \cdot \frac{\pi}{6} - 0 = \frac{\pi}{12}$ . For the integral, $w = 1 - x^2$ , $dw = -2x\,dx$ , so an antiderivative of $\frac{x}{\sqrt{1 - x^2}}$ is $-\sqrt{1 - x^2}$ , and $\int_0^{1/2} \frac{x}{\sqrt{1 - x^2}}\,dx = \left[-\sqrt{1 - x^2}\right]_0^{1/2} = -\frac{\sqrt 3}{2} + 1$ . Hence $\int_0^{1/2} \arcsin x\,dx = \frac{\pi}{12} - 1 + \frac{\sqrt 3}{2}$ . Plausibility, with $\pi \approx 3.14$ and $\sqrt 3 \approx 1.73$ : $0.26 + 0.87 - 1 \approx 0.13$ . The graph of $\arcsin$ is convex on $[0, \frac{1}{2}]$ , so it lies below its chord from $(0, 0)$ to $(\frac{1}{2}, \frac{\pi}{6})$ , and the area must be less than that triangle, $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{24} \approx 0.131$ . It is, just barely, as the near-straight graph suggests.

d) $\ln x \ge 0$ on $[1, e]$ , so the area is the integral: $A = \int_1^e \ln x\,dx = \left[x \ln x - x\right]_1^e = (e \cdot 1 - e) - (1 \cdot 0 - 1) = 0 + 1 = 1$ . Plausibility: the region sits inside the rectangle $[1, e] \times [0, 1]$ of area $e - 1 \approx 1.72$ , and since $\ln$ is concave its graph lies ABOVE the chord from $(1, 0)$ to $(e, 1)$ , so $A$ exceeds the triangle of area $\frac{e - 1}{2} \approx 0.86$ . The value $1$ sits between the two. A result like $e - 1$ or $e$ (the typical outcome of dropping the $-x$ ) is ruled out by the figure at once.

e) $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$ : that student DIFFERENTIATED $\ln x$ instead of integrating it. $\frac{d}{dx}\frac{(\ln x)^2}{2} = \frac{\ln x}{x}$ : the power rule was applied as if $\ln x$ were the variable, forgetting the chain rule factor $\frac{1}{x}$ ; that is the antiderivative of $\frac{\ln x}{x}$ , a substitution problem, not of $\ln x$ . $\frac{d}{dx}(x \ln x) = \ln x + 1$ : this student wrote $uv$ and stopped, forgetting the trade's second half, $-\int v\,du = -x$ . All three errors are caught by the same thirty-second derivative, which is why it belongs at the end of every answer.

Exercise 3: Repeated parts and the tabular method

When $u$ is a polynomial of degree $n$ , each integration by parts lowers its degree by one, so $n$ integrations finish the job. The tabular method organizes them: a column D of successive derivatives of $u$ down to $0$ , a column I of successive antiderivatives of $dv$ , and signs that ALTERNATE, $+, -, +, -, \dots$ , starting with $+$ .

The table is only a bookkeeping of repeated parts: every line of it is one trade, and every minus sign is the minus of $-\int v\,du$ .

a) Compute $\int x^2 e^{-x}\,dx$ by two explicit integrations by parts.
b) Compute $\int x^3 \cos(2x)\,dx$ with the tabular method, showing the table.
c) Compute $\int_0^{\pi} x^2 \cos x\,dx$ and explain the sign of the result.
d) Explain why the table is useless for $\int x^2 \ln x\,dx$ and for $\int e^x \sin x\,dx$ , and say what to do instead for the first one.
e) A student uses the table for $\int x^2 e^{-x}\,dx$ with the signs $+, +, +$ and gets $-e^{-x}(x^2 - 2x + 2) + C$ . Differentiate his answer and locate the error.

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Answers

a) $\int x^2 e^{-x}\,dx = -e^{-x}(x^2 + 2x + 2) + C$
b) $\frac{x^3 \sin 2x}{2} + \frac{3x^2 \cos 2x}{4} - \frac{3x \sin 2x}{4} - \frac{3\cos 2x}{8} + C$
c) $\int_0^{\pi} x^2 \cos x\,dx = -2\pi$
d) The D column of $\ln x$ never reaches $0$ ; for $e^x \sin x$ neither column ends. $\int x^2 \ln x\,dx = \frac{x^3 \ln x}{3} - \frac{x^3}{9} + C$ by one trade.
e) His derivative is $e^{-x}(x - 2)^2$ ; the signs must alternate.

a) First trade: $u = x^2$ , $du = 2x\,dx$ ; $dv = e^{-x}\,dx$ , $v = -e^{-x}$ . So $\int x^2 e^{-x}\,dx = -x^2 e^{-x} + \int 2x e^{-x}\,dx$ : the degree went from $2$ to $1$ , the trade is good. Second trade on the new integral: $u = 2x$ , $du = 2\,dx$ ; $dv = e^{-x}\,dx$ , $v = -e^{-x}$ , so $\int 2x e^{-x}\,dx = -2x e^{-x} + \int 2e^{-x}\,dx = -2x e^{-x} - 2e^{-x}$ . Total: $-x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + C = -e^{-x}(x^2 + 2x + 2) + C$ . Check: $e^{-x}(x^2 + 2x + 2) - e^{-x}(2x + 2) = x^2 e^{-x}$ . The sign of $v = -e^{-x}$ is where marks go: it enters twice, and $-(-e^{-x})$ becomes a PLUS in front of the new integral.

b) D column: $x^3$ , $3x^2$ , $6x$ , $6$ , $0$ . I column: $\cos 2x$ , $\frac{\sin 2x}{2}$ , $-\frac{\cos 2x}{4}$ , $-\frac{\sin 2x}{8}$ , $\frac{\cos 2x}{16}$ . Multiply each D entry by the I entry ONE ROW BELOW, with the signs $+, -, +, -$ : $+x^3 \cdot \frac{\sin 2x}{2}$ , $-3x^2 \cdot \left(-\frac{\cos 2x}{4}\right)$ , $+6x \cdot \left(-\frac{\sin 2x}{8}\right)$ , $-6 \cdot \frac{\cos 2x}{16}$ . So $\int x^3 \cos 2x\,dx = \frac{x^3 \sin 2x}{2} + \frac{3x^2 \cos 2x}{4} - \frac{3x \sin 2x}{4} - \frac{3\cos 2x}{8} + C$ . Three integrations by parts would have given the same four terms in three times the space. The diagonal pattern, D in row $k$ with I in row $k + 1$ , is the part to get right: pairing each D with the I of the SAME row gives the product $x^3 \cos 2x$ back, which is the integrand, not its antiderivative.

c) Table: D $x^2$ , $2x$ , $2$ , $0$ ; I $\cos x$ , $\sin x$ , $-\cos x$ , $-\sin x$ . Antiderivative: $x^2 \sin x + 2x \cos x - 2\sin x$ . At $\pi$ : $0 + 2\pi(-1) - 0 = -2\pi$ ; at $0$ : $0$ . So $\int_0^{\pi} x^2 \cos x\,dx = -2\pi$ . The sign is right: $\cos x > 0$ on $(0, \frac{\pi}{2})$ and $< 0$ on $(\frac{\pi}{2}, \pi)$ , but the weight $x^2$ is much larger on the second half, so the negative part wins. A positive answer here would be a signal to recheck the table's signs, not a result to report.

d) For $\int x^2 \ln x\,dx$ , putting $\ln x$ in the D column gives $\frac{1}{x}$ , $-\frac{1}{x^2}$ , $\frac{2}{x^3}$ , and so on: it never reaches $0$ . Putting $x^2$ in D instead requires antiderivatives of $\ln x$ , then of $x \ln x - x$ , and so on: worse at every row. The right move is ONE trade: $u = \ln x$ , $dv = x^2\,dx$ , $v = \frac{x^3}{3}$ , giving $\frac{x^3 \ln x}{3} - \int \frac{x^2}{3}\,dx = \frac{x^3 \ln x}{3} - \frac{x^3}{9} + C$ . For $\int e^x \sin x\,dx$ , the derivatives of either factor cycle forever ( $\sin$ , $\cos$ , $-\sin$ , $\dots$ or $e^x$ , $e^x$ , $\dots$ ): no column reaches $0$ . That integral is solved by making the original integral come back, which is the next exercise. The table is a tool for ONE form only: polynomial times a function you can integrate repeatedly.

e) $\frac{d}{dx}\left[-e^{-x}(x^2 - 2x + 2)\right] = e^{-x}(x^2 - 2x + 2) - e^{-x}(2x - 2) = e^{-x}(x^2 - 4x + 4) = e^{-x}(x - 2)^2$ , which is not $x^2 e^{-x}$ . With the D column $x^2$ , $2x$ , $2$ and the I column $e^{-x}$ , $-e^{-x}$ , $e^{-x}$ , $-e^{-x}$ , his signs give $x^2(-e^{-x}) + 2x(e^{-x}) + 2(-e^{-x})$ ; the correct alternation $+, -, +$ flips the middle term to $-2x e^{-x}$ , which yields the answer of a). The minus signs of the table are the minus of $-\int v\,du$ , applied once per trade, so they cannot all be dropped. The derivative check located the error without redoing the table.

Exercise 4: Cyclic parts: the integral that comes back

For $\int e^x \sin x\,dx$ no factor simplifies when differentiated: $\sin$ becomes $\cos$ , then $-\sin$ , and $e^x$ stays $e^x$ . Two integrations by parts bring back the ORIGINAL integral, with a coefficient. You then stop integrating and solve for it, like an unknown in an equation.

The figure shows the graph of $y = e^x \sin x$ on $[0, \pi]$ and the region $A$ under it.

a) Let $I = \int e^x \sin x\,dx$ . Integrate by parts twice with $u = \sin x$ , then $u = \cos x$ , and solve for $I$ .
b) A student does the first step as in a), then in the second step takes $u = e^x$ and $dv = \cos x\,dx$ . Show what he obtains, and state the rule that avoids it.
c) Compute the area of the region $A$ exactly, and check its size against the figure.
d) Compute $\int e^{2x} \cos 3x\,dx$ .
e) Compute $\int \sin(\ln x)\,dx$ , for $x > 0$ , with $dv = dx$ in both steps.

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Answers

a) $\int e^x \sin x\,dx = \frac{e^x(\sin x - \cos x)}{2} + C$
b) He gets $I = I$ , true and empty; keep the SAME type of function as $u$ in both steps.
c) $A = \frac{e^{\pi} + 1}{2}$ , more than $10$ .
d) $\int e^{2x} \cos 3x\,dx = \frac{e^{2x}(2\cos 3x + 3\sin 3x)}{13} + C$
e) $\int \sin(\ln x)\,dx = \frac{x\left(\sin(\ln x) - \cos(\ln x)\right)}{2} + C$

a) First trade: $u = \sin x$ , $du = \cos x\,dx$ ; $dv = e^x\,dx$ , $v = e^x$ . So $I = e^x \sin x - \int e^x \cos x\,dx$ . Second trade, on the new integral: $u = \cos x$ , $du = -\sin x\,dx$ ; $dv = e^x\,dx$ , $v = e^x$ . So $\int e^x \cos x\,dx = e^x \cos x + \int e^x \sin x\,dx = e^x \cos x + I$ . Substituting: $I = e^x \sin x - e^x \cos x - I$ . The integral came back with coefficient $-1$ , so move it to the left: $2I = e^x(\sin x - \cos x)$ , and $I = \frac{e^x(\sin x - \cos x)}{2} + C$ . The constant is added at the END, after dividing: writing $+C$ inside the equation and then halving it changes nothing, but forgetting to divide the WHOLE right side by $2$ is the classic loss. Check: $\frac{1}{2}\left[e^x(\sin x - \cos x) + e^x(\cos x + \sin x)\right] = e^x \sin x$ .

b) Second step with $u = e^x$ , $du = e^x\,dx$ ; $dv = \cos x\,dx$ , $v = \sin x$ : $\int e^x \cos x\,dx = e^x \sin x - \int e^x \sin x\,dx = e^x \sin x - I$ . Substituting into the first step: $I = e^x \sin x - (e^x \sin x - I) = I$ . The equation is true and says nothing: the second trade simply UNDID the first, handing back the integral it had produced. The rule: in a cyclic integral, whichever type you chose as $u$ in the first step (here the trigonometric factor), keep the same type as $u$ in the second. Both consistent choices work, $u$ trigonometric twice or $u$ exponential twice; only the mixed one goes in a circle.

c) $e^x > 0$ and $\sin x \ge 0$ on $[0, \pi]$ , so $A = \int_0^{\pi} e^x \sin x\,dx = \left[\frac{e^x(\sin x - \cos x)}{2}\right]_0^{\pi} = \frac{e^{\pi}(0 + 1)}{2} - \frac{1 \cdot (0 - 1)}{2} = \frac{e^{\pi} + 1}{2}$ . At the upper bound $\cos \pi = -1$ turns $-\cos x$ into $+1$ , and at the lower bound $-\cos 0 = -1$ , SUBTRACTED, gives the second $+\frac{1}{2}$ . Size, without a calculator: $e^3 \approx 20$ (since $e^3 = e \cdot e^2 \approx 2.72 \cdot 7.4$ ), so $e^{\pi} > 20$ and $A > \frac{21}{2}$ , more than $10$ . On the figure, the curve rises to about $7.5$ near $x = \frac{3\pi}{4}$ over a base of length $\pi$ ; a triangle with that height and base has area about $11.7$ , the same order. A value like $\frac{e^{\pi} - 1}{2}$ , from a sign slip at $x = 0$ , would be within the same range, which is why the bracket must be written out at BOTH bounds.

d) Let $J = \int e^{2x} \cos 3x\,dx$ . First trade: $u = \cos 3x$ , $du = -3\sin 3x\,dx$ ; $dv = e^{2x}\,dx$ , $v = \frac{e^{2x}}{2}$ : $J = \frac{e^{2x}\cos 3x}{2} + \frac{3}{2}\int e^{2x} \sin 3x\,dx$ . Second trade, SAME type as $u$ : $u = \sin 3x$ , $du = 3\cos 3x\,dx$ , $v = \frac{e^{2x}}{2}$ : $\int e^{2x} \sin 3x\,dx = \frac{e^{2x}\sin 3x}{2} - \frac{3}{2}J$ . So $J = \frac{e^{2x}\cos 3x}{2} + \frac{3e^{2x}\sin 3x}{4} - \frac{9}{4}J$ , that is $\frac{13}{4}J = \frac{e^{2x}(2\cos 3x + 3\sin 3x)}{4}$ and $J = \frac{e^{2x}(2\cos 3x + 3\sin 3x)}{13} + C$ . The coefficient $1 + \frac{9}{4} = \frac{13}{4}$ is where the work is: the number $13 = 2^2 + 3^2$ always appears for $e^{ax}$ with $\cos bx$ , which is a free check on the arithmetic.

e) Let $K = \int \sin(\ln x)\,dx$ . First trade: $u = \sin(\ln x)$ , $du = \frac{\cos(\ln x)}{x}\,dx$ ; $dv = dx$ , $v = x$ : $K = x\sin(\ln x) - \int \cos(\ln x)\,dx$ , the $x$ of $v$ cancelling the $\frac{1}{x}$ of the chain rule. Second trade, same type: $u = \cos(\ln x)$ , $du = -\frac{\sin(\ln x)}{x}\,dx$ , $v = x$ : $\int \cos(\ln x)\,dx = x\cos(\ln x) + K$ . Hence $K = x\sin(\ln x) - x\cos(\ln x) - K$ and $K = \frac{x\left(\sin(\ln x) - \cos(\ln x)\right)}{2} + C$ . This is the lone-factor gesture of exercise 2 and the cyclic gesture of this exercise at once: nothing in the integrand looks like a product, yet parts with $dv = dx$ produces the cycle.

Exercise 5: Parts on a definite integral: the bracket is evaluated at both bounds

On a definite integral the trade reads $\int_a^b u\,dv = \left[uv\right]_a^b - \int_a^b v\,du$ . The bracket is a NUMBER, $u(b)v(b) - u(a)v(a)$ , and the new integral keeps the same bounds. The two halves are evaluated separately, and each one at both ends.

The figure shows the graph of $y = \arcsin x$ for $0 \le x \le \frac{1}{2}$ , the region $A$ under it and the region $B$ between the curve and the $y$ -axis, inside the rectangle of corners $(0, 0)$ and $(\frac{1}{2}, \frac{\pi}{6})$ .

a) Compute $\int_1^e x \ln x\,dx$ .
b) Compute $\int_0^{\pi/2} x \cos x\,dx$ .
c) Compute $\int_0^{\pi} x \sin x\,dx$ . Which half of the formula carries the value here?
d) Using the figure, express the area of $B$ as an integral in $y$ , then recover $\int_0^{1/2} \arcsin x\,dx$ from $A + B$ without integrating by parts. Which term of the parts formula is the rectangle?
e) A student finds $\int_1^2 \ln x\,dx = 2\ln 2 - 2$ . Find the correct value and explain how a size check rules out his answer before any recomputation.

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Answers

a) $\int_1^e x \ln x\,dx = \frac{e^2 + 1}{4}$
b) $\int_0^{\pi/2} x \cos x\,dx = \frac{\pi}{2} - 1$
c) $\int_0^{\pi} x \sin x\,dx = \pi$ , all of it from the bracket.
d) $B = \int_0^{\pi/6} \sin y\,dy = 1 - \frac{\sqrt 3}{2}$ , so $A = \frac{\pi}{12} + \frac{\sqrt 3}{2} - 1$ ; the rectangle is $\left[x \arcsin x\right]_0^{1/2}$ .
e) $\int_1^2 \ln x\,dx = 2\ln 2 - 1$ ; his value is negative for a positive integrand.

a) $u = \ln x$ , $du = \frac{dx}{x}$ ; $dv = x\,dx$ , $v = \frac{x^2}{2}$ . Then $\int_1^e x \ln x\,dx = \left[\frac{x^2 \ln x}{2}\right]_1^e - \int_1^e \frac{x}{2}\,dx$ . The bracket is $\frac{e^2 \cdot 1}{2} - \frac{1 \cdot 0}{2} = \frac{e^2}{2}$ . The new integral is $\left[\frac{x^2}{4}\right]_1^e = \frac{e^2 - 1}{4}$ . So the value is $\frac{e^2}{2} - \frac{e^2 - 1}{4} = \frac{e^2 + 1}{4}$ . With $e^2 \approx 7.4$ that is about $2.1$ ; the integrand lies between $0$ and $e$ on an interval of length $e - 1 \approx 1.7$ , so the value must be below $e(e - 1) \approx 4.7$ , which it is. The lower bound contributed $0$ to the bracket because $\ln 1 = 0$ , but NOT to the new integral, where $\frac{1}{4}$ survives: that $\frac{1}{4}$ is exactly what gets lost when the lower bound is evaluated only once.

b) $u = x$ , $du = dx$ ; $dv = \cos x\,dx$ , $v = \sin x$ . Then $\int_0^{\pi/2} x \cos x\,dx = \left[x \sin x\right]_0^{\pi/2} - \int_0^{\pi/2} \sin x\,dx = \frac{\pi}{2} \cdot 1 - 0 - \left[-\cos x\right]_0^{\pi/2} = \frac{\pi}{2} - (0 + 1) = \frac{\pi}{2} - 1$ . With $\pi \approx 3.14$ , about $0.57$ : positive, as it must be since both factors are non-negative on $[0, \frac{\pi}{2}]$ . The step $\left[-\cos x\right]_0^{\pi/2} = -\cos\frac{\pi}{2} + \cos 0 = 1$ carries two signs, and writing it on its own line is how they survive.

c) $u = x$ , $du = dx$ ; $dv = \sin x\,dx$ , $v = -\cos x$ . Then $\int_0^{\pi} x \sin x\,dx = \left[-x \cos x\right]_0^{\pi} + \int_0^{\pi} \cos x\,dx = (-\pi \cdot (-1) - 0) + \left[\sin x\right]_0^{\pi} = \pi + 0 = \pi$ . Here the new integral contributes NOTHING and the whole value is the bracket, the opposite of part b), where the bracket was $\frac{\pi}{2}$ and the new integral took $1$ away. Neither half can be skipped in advance: a student who assumes the bracket vanishes (it often does at $x = 0$ ) would answer $0$ here, for the area of a region that is visibly positive.

d) Slice $B$ horizontally: at height $y$ , with $0 \le y \le \frac{\pi}{6}$ , the region runs from the $y$ -axis to the curve, where $y = \arcsin x$ , that is $x = \sin y$ . So $B = \int_0^{\pi/6} \sin y\,dy = \left[-\cos y\right]_0^{\pi/6} = 1 - \frac{\sqrt 3}{2}$ . The two regions fill the rectangle: $A + B = \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12}$ . Hence $A = \int_0^{1/2} \arcsin x\,dx = \frac{\pi}{12} - 1 + \frac{\sqrt 3}{2}$ , the value found by parts in exercise 2. The rectangle is the bracket $\left[x \arcsin x\right]_0^{1/2}$ , and $B$ is the new integral $\int x\,du$ with $u = \arcsin x$ , measured along the vertical axis. That is what integration by parts says geometrically: $\int u\,dv + \int v\,du$ is the difference of two rectangles, $u(b)v(b) - u(a)v(a)$ .

e) $\int_1^2 \ln x\,dx = \left[x \ln x - x\right]_1^2 = (2\ln 2 - 2) - (1 \cdot 0 - 1) = 2\ln 2 - 1$ . The student evaluated the antiderivative at the upper bound only, reasoning that $\ln 1 = 0$ kills the lower bound; it kills $x \ln x$ but not $-x$ , which contributes $+1$ . Size check: on $[1, 2]$ , $0 \le \ln x \le \ln 2$ , so the integral lies between $0$ and $\ln 2 \approx 0.69$ . His value $2\ln 2 - 2 \approx -0.61$ is NEGATIVE for a non-negative integrand: impossible, and visible in five seconds. The correct $2\ln 2 - 1 \approx 0.39$ lies in the window.

Exercise 6: A reduction formula: the trade written once for every n

When the same integration by parts would be repeated $n$ times, it is written ONCE with a letter, and the result links the integral of index $n$ to the one of index $n - 1$ : a reduction formula. For $n \ge 0$ , let $I_n = \int_0^1 x^n e^x\,dx$ .

The figure shows the four integrands $x^n e^x$ for $n = 0, 1, 2, 3$ on $[0, 1]$ ; from top to bottom, the curves are in the same order as the list on the left. They all meet at $(1, e)$ .

a) Prove that $I_n = e - n I_{n-1}$ for every $n \ge 1$ . Where is the hypothesis $n \ge 1$ used?
b) Compute $I_0$ , then $I_1$ , $I_2$ and $I_3$ exactly.
c) Prove that $\frac{1}{n + 1} \le I_n \le \frac{e}{n + 1}$ , and check $I_2$ and $I_3$ against these bounds. What does the figure say about the order of the $I_n$ ?
d) Let $J_n = \int_1^e (\ln x)^n\,dx$ . Prove that $J_n = e - n J_{n-1}$ for $n \ge 1$ , and deduce $J_2$ and $J_3$ without any new integration. Explain the coincidence with a substitution.
e) A student starts the recursion with $I_0 = e$ . Compute what he obtains for $I_3$ , and explain how an error in $I_0$ grows along the recursion.

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a) $u = x^n$ , $dv = e^x\,dx$ : $I_n = \left[x^n e^x\right]_0^1 - n I_{n-1} = e - n I_{n-1}$ ; $n \ge 1$ makes $0^n = 0$ .
b) $I_0 = e - 1$ , $I_1 = 1$ , $I_2 = e - 2$ , $I_3 = 6 - 2e$
c) $1 \le e^x \le e$ on $[0, 1]$ ; $\frac{1}{3} \le e - 2 \le \frac{e}{3}$ and $\frac{1}{4} \le 6 - 2e \le \frac{e}{4}$ ; $I_0 > I_1 > I_2 > I_3$ .
d) $J_n = I_n$ for all $n$ : $J_2 = e - 2$ , $J_3 = 6 - 2e$ (substitution $x = e^t$ ).
e) He gets $I_3 = -2e < 0$ ; an error $\varepsilon$ in $I_0$ becomes $(-1)^n n!\,\varepsilon$ in $I_n$ .

a) Integrate by parts with $u = x^n$ , $du = n x^{n-1}\,dx$ ; $dv = e^x\,dx$ , $v = e^x$ . Then $I_n = \left[x^n e^x\right]_0^1 - \int_0^1 n x^{n-1} e^x\,dx = (1^n e^1 - 0^n e^0) - n I_{n-1}$ . For $n \ge 1$ , $0^n = 0$ , so the bracket is $e$ and $I_n = e - n I_{n-1}$ . The hypothesis is used exactly there: for $n = 0$ the lower bound would give $0^0 e^0 = 1$ , and the formula would not hold (it would not even make sense, since $I_{-1}$ is not defined). The choice $u = x^n$ is the one of exercise 1, and it is right for the same reason: the power goes DOWN, from $n$ to $n - 1$ , which is what makes the formula a reduction.

b) $I_0 = \int_0^1 e^x\,dx = e - 1$ . Then $I_1 = e - 1 \cdot I_0 = e - (e - 1) = 1$ , $I_2 = e - 2I_1 = e - 2$ , $I_3 = e - 3I_2 = e - 3(e - 2) = 6 - 2e$ . Each value costs one line; computing $I_3$ by the tabular method would give the antiderivative $e^x(x^3 - 3x^2 + 6x - 6)$ , whose value at $1$ minus its value at $0$ is $-2e + 6$ , the same number. Size: with $e \approx 2.72$ , $I_3 \approx 6 - 5.44 = 0.56$ .

c) On $[0, 1]$ , $1 \le e^x \le e$ , so $x^n \le x^n e^x \le e\,x^n$ , and integrating (the comparison property of the definite integral) gives $\frac{1}{n + 1} \le I_n \le \frac{e}{n + 1}$ , since $\int_0^1 x^n\,dx = \frac{1}{n + 1}$ . For $n = 2$ : $\frac{1}{3} \approx 0.33 \le e - 2 \approx 0.72 \le \frac{e}{3} \approx 0.91$ . For $n = 3$ : $\frac{1}{4} = 0.25 \le 6 - 2e \approx 0.56 \le \frac{e}{4} \approx 0.68$ . Both hold. On the figure, for $0 < x < 1$ , $x^{n+1} < x^n$ , so each curve lies BELOW the previous one: the areas decrease, $I_0 > I_1 > I_2 > I_3 > 0$ , and the bounds show they tend to $0$ . These inequalities are the check that the recursion was run correctly, and they cost nothing.

d) $u = (\ln x)^n$ , $du = n(\ln x)^{n-1} \cdot \frac{1}{x}\,dx$ ; $dv = dx$ , $v = x$ . Then $J_n = \left[x (\ln x)^n\right]_1^e - \int_1^e x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x}\,dx = (e \cdot 1 - 1 \cdot 0) - n J_{n-1} = e - n J_{n-1}$ , where $(\ln 1)^n = 0$ uses $n \ge 1$ again. The start is $J_0 = \int_1^e dx = e - 1 = I_0$ . Same recursion, same first term: $J_n = I_n$ for every $n$ , so $J_2 = e - 2$ and $J_3 = 6 - 2e$ with no new integration. The coincidence is a substitution: $x = e^t$ , $dx = e^t\,dt$ , with $t$ from $0$ to $1$ , turns $J_n$ into $\int_0^1 t^n e^t\,dt = I_n$ . The lone-logarithm trade of exercise 2 and the power-lowering trade of exercise 1 are the SAME trade, seen in two variables.

e) With $I_0 = e$ : $I_1 = e - e = 0$ , $I_2 = e - 0 = e$ , $I_3 = e - 3e = -2e$ . A negative value for the integral of a positive function: impossible, and part c) catches it at once. The mechanism: if the value used for $I_{n-1}$ is off by $\varepsilon$ , then $I_n = e - nI_{n-1}$ is off by $-n\varepsilon$ . Starting from an error $\varepsilon = 1$ in $I_0$ , the errors are $-1$ , $+2$ , $-6$ in $I_1$ , $I_2$ , $I_3$ : the error is multiplied by $(-1)^n n!$ . A reduction formula is exact, but it AMPLIFIES any slip in the first term, so $I_0$ is the value to double-check.

Exercise 7: Parts when the function is unknown: moving the derivative

Integration by parts does not need a formula for the functions involved. It moves a derivative from one factor to the other, and the bracket only needs VALUES at the bounds. This is how exam questions test the method on functions known only through a table.

A function $f$ has a continuous second derivative on $[1, 4]$ , and $f(1) = 2$ , $f(4) = 7$ , $f'(1) = 5$ , $f'(4) = 3$ . Moreover $\int_1^4 f(x)\,dx = 12$ .

a) Compute $\int_1^4 x f''(x)\,dx$ .
b) Explain why the choice $u = f''(x)$ , $dv = x\,dx$ is not available here.
c) Compute $\int_1^4 x f'(x)\,dx$ .
d) Let $F(x) = \int_x^1 e^{-t^2}\,dt$ . Compute $\int_0^1 F(x)\,dx$ exactly, by integrating by parts with $u = F(x)$ .
e) A student gets $-\frac{1 - e^{-1}}{2}$ in d). Locate the error, and give bounds for $\int_0^1 F(x)\,dx$ that rule out his sign.

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Answers

a) $\int_1^4 x f''(x)\,dx = \left[x f'(x)\right]_1^4 - (f(4) - f(1)) = 7 - 5 = 2$
b) It needs $f'''$ , which is not given and not even assumed to exist, and it raises the power of $x$ .
c) $\int_1^4 x f'(x)\,dx = 28 - 2 - 12 = 14$
d) $\int_0^1 F(x)\,dx = \frac{1 - e^{-1}}{2}$
e) He wrote $F'(x) = +e^{-x^2}$ ; the value lies between $\frac{e^{-1}}{2}$ and $\frac{1}{2}$ , so it is positive.

a) $u = x$ , $du = dx$ ; $dv = f''(x)\,dx$ , $v = f'(x)$ . Then $\int_1^4 x f''(x)\,dx = \left[x f'(x)\right]_1^4 - \int_1^4 f'(x)\,dx$ . The bracket needs only values: $4f'(4) - 1 \cdot f'(1) = 12 - 5 = 7$ . The new integral is a net change (Fundamental Theorem): $\int_1^4 f'(x)\,dx = f(4) - f(1) = 7 - 2 = 5$ . So the value is $7 - 5 = 2$ . The derivative has moved from $f''$ onto the factor $x$ , where it became $1$ , and what remained, $f'$ , has an antiderivative we can evaluate: $f$ .

b) With $u = f''(x)$ we would need $du = f'''(x)\,dx$ , and the statement only guarantees that $f''$ is continuous: $f'''$ may not exist, and even if it did, nothing is known about it. The new integral would be $\int_1^4 \frac{x^2}{2} f'''(x)\,dx$ , a higher power AND a higher derivative, worse on both counts. The general principle of the chapter applies to unknown functions too: put the derivative where it simplifies ( $x$ becomes $1$ ) and take as $dv$ what you can integrate ( $f''$ integrates to $f'$ ).

c) $u = x$ , $du = dx$ ; $dv = f'(x)\,dx$ , $v = f(x)$ . Then $\int_1^4 x f'(x)\,dx = \left[x f(x)\right]_1^4 - \int_1^4 f(x)\,dx = (4 \cdot 7 - 1 \cdot 2) - 12 = 28 - 2 - 12 = 14$ . This time the given value of $\int_1^4 f(x)\,dx$ is indispensable: without it the question could not be answered, which is why it is in the statement. The data are consistent: a polynomial of degree $4$ satisfies all five conditions, so the numbers describe a real function.

d) $u = F(x)$ , $du = F'(x)\,dx$ ; $dv = dx$ , $v = x$ . By the Fundamental Theorem, part 1, with the variable at the LOWER bound, $F'(x) = -e^{-x^2}$ . So $\int_0^1 F(x)\,dx = \left[x F(x)\right]_0^1 - \int_0^1 x \cdot (-e^{-x^2})\,dx$ . The bracket vanishes at both ends: $F(1) = \int_1^1 e^{-t^2}\,dt = 0$ , and at $x = 0$ the factor $x$ is $0$ . What remains is $\int_0^1 x e^{-x^2}\,dx$ , a substitution ( $w = x^2$ , $dw = 2x\,dx$ ): $\left[-\frac{1}{2}e^{-x^2}\right]_0^1 = \frac{1 - e^{-1}}{2}$ . The integrand $e^{-t^2}$ has no elementary antiderivative, so $F$ has no formula; parts sidesteps that entirely, because it only uses $F'$ , which the Fundamental Theorem gives for free. That is the lone-factor gesture of exercise 2 again, with $dv = dx$ .

e) His value is the correct one with the opposite sign: he took $F'(x) = e^{-x^2}$ , forgetting that the variable is the LOWER bound, which contributes a minus sign. Bounds without computing: for $0 \le t \le 1$ , $e^{-1} \le e^{-t^2} \le 1$ , so for $0 \le x \le 1$ , $e^{-1}(1 - x) \le F(x) \le 1 - x$ . Integrating over $[0, 1]$ : $\frac{e^{-1}}{2} \le \int_0^1 F(x)\,dx \le \frac{1}{2}$ . The integral is positive, about between $0.18$ and $0.5$ , and $\frac{1 - e^{-1}}{2} \approx \frac{1 - 0.37}{2} \approx 0.32$ sits inside. A negative answer for the integral of a function that is $\ge 0$ is always a sign error, and here it points straight at the derivative of an integral with a variable lower bound.

Exercise 8: Five statements to correct

Each statement below was written on a real exam paper. Say whether it is true or false; when it is false, find the error, correct it, and prove the correction with a derivative or a number.

Two marks per statement: one for locating the error, one for the corrected result with its check.

a) $\int x e^x\,dx = \left(\int x\,dx\right)\left(\int e^x\,dx\right) = \frac{x^2}{2}e^x + C$ .
b) $\int x \sin x\,dx = -x\cos x - \sin x + C$ .
c) By the tabular method, $\int x^2 e^{3x}\,dx = e^{3x}\left(\frac{x^2}{3} + \frac{2x}{9} + \frac{2}{27}\right) + C$ .
d) $\int_0^1 x e^x\,dx = \left[x e^x\right]_0^1 - \int e^x\,dx = e - e^x + C$ .
e) When $v$ is computed from $dv$ , its constant must be taken equal to $0$ , otherwise integration by parts gives a wrong result.

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Answers

a) False: $\int x e^x\,dx = (x - 1)e^x + C$ .
b) False: $\int x \sin x\,dx = -x\cos x + \sin x + C$ .
c) False: $e^{3x}\left(\frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27}\right) + C$ .
d) False: $\int_0^1 x e^x\,dx = e - (e - 1) = 1$ .
e) False: any constant works; $v = x + 1$ gives $\int \ln(x + 1)\,dx = (x + 1)\ln(x + 1) - x + C$ in one line.

a) False. There is no product rule for integrals: the integral of a product is not the product of the integrals. The derivative of the proposed answer is $x e^x + \frac{x^2}{2}e^x \ne x e^x$ . Correct method: $u = x$ , $dv = e^x\,dx$ , $v = e^x$ , so $\int x e^x\,dx = x e^x - \int e^x\,dx = (x - 1)e^x + C$ ; check, $e^x + (x - 1)e^x = x e^x$ . On $[0, 1]$ the difference is visible in numbers: the correct integral is $1$ , while $\left(\int_0^1 x\,dx\right)\left(\int_0^1 e^x\,dx\right) = \frac{e - 1}{2} \approx 0.86$ . Integration by parts EXISTS precisely because the naive rule is false.

b) False, one sign. $u = x$ , $dv = \sin x\,dx$ , $v = -\cos x$ : $\int x\sin x\,dx = -x\cos x - \int (-\cos x)\,dx = -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C$ . The student subtracted $\int v\,du$ without carrying the minus sign INSIDE $v$ . The derivative of his answer is $-\cos x + x\sin x - \cos x = x\sin x - 2\cos x$ , and the stray $-2\cos x$ is the trace of the lost sign; the corrected answer differentiates to $-\cos x + x\sin x + \cos x = x\sin x$ . Writing $v = -\cos x$ in its own box, then $-\int v\,du = +\int \cos x\,dx$ on a separate line, prevents it.

c) False: the signs of the table must alternate. D column: $x^2$ , $2x$ , $2$ , $0$ ; I column: $e^{3x}$ , $\frac{e^{3x}}{3}$ , $\frac{e^{3x}}{9}$ , $\frac{e^{3x}}{27}$ . With the signs $+, -, +$ : $\frac{x^2 e^{3x}}{3} - \frac{2x e^{3x}}{9} + \frac{2e^{3x}}{27}$ , so $\int x^2 e^{3x}\,dx = e^{3x}\left(\frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27}\right) + C$ . Check: $3e^{3x}\left(\frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27}\right) + e^{3x}\left(\frac{2x}{3} - \frac{2}{9}\right) = e^{3x}\left(x^2 - \frac{2x}{3} + \frac{2}{9} + \frac{2x}{3} - \frac{2}{9}\right) = x^2 e^{3x}$ . The terms cancel in pairs only with the alternating signs; with all plus signs the derivative is $e^{3x}\left(x^2 + \frac{4x}{3} + \frac{4}{9}\right)$ .

d) False: a definite integral is a number, and this answer is a function plus a constant. The bracket was evaluated, but the new integral was treated as indefinite. Correct: $\int_0^1 x e^x\,dx = \left[x e^x\right]_0^1 - \int_0^1 e^x\,dx = (e - 0) - (e - 1) = 1$ . The bounds stay on BOTH halves of the formula. A quick sanity check: $0 \le x e^x \le e$ on $[0, 1]$ , so the value lies between $0$ and $e$ , and $1$ does.

e) False. If $v$ is replaced by $v + K$ , the formula gives $u(v + K) - \int (v + K)\,du = uv - \int v\,du + Ku - Ku + \text{const}$ , the same family of antiderivatives. So any constant is allowed, and a clever one can shorten the work. For $\int \ln(x + 1)\,dx$ : $u = \ln(x + 1)$ , $du = \frac{dx}{x + 1}$ , $dv = dx$ , and choosing $v = x + 1$ gives $(x + 1)\ln(x + 1) - \int 1\,dx = (x + 1)\ln(x + 1) - x + C$ at once. With $v = x$ one gets $x\ln(x + 1) - \int \frac{x}{x + 1}\,dx = x\ln(x + 1) - x + \ln(x + 1) + C$ , after writing $\frac{x}{x + 1} = 1 - \frac{1}{x + 1}$ : the same function, one line longer. Check: $\ln(x + 1) + 1 - 1 = \ln(x + 1)$ .

Exercise 9: A damped oscillation: how far does the mass really travel?

A mass on a spring, slowed by friction, moves along a horizontal line. Its velocity, in metres per second, is $v(t) = e^{-t}\sin(\pi t)$ for $t \ge 0$ in seconds. The factor $\sin(\pi t)$ makes it change direction every second; the factor $e^{-t}$ damps the motion.

The figure shows $v$ on $[0, 2]$ : the first shaded lobe is the first swing, the second lobe, below the axis, is the swing back. Displacement is the integral of $v$ ; distance travelled is the integral of $|v|$ .

a) Find an antiderivative $P$ of $v$ by cyclic integration by parts.
b) Compute the displacement during the first second, $\int_0^1 v(t)\,dt$ .
c) Compute the displacement during the next second, and the ratio of the two displacements.
d) Give the net displacement and the total distance travelled over $[0, 2]$ .
e) Show that the displacement during $[k, k + 1]$ is $(-1)^k e^{-k}$ times the first one. With $\pi \approx 3.14$ and $e \approx 2.72$ , estimate the first swing, and find the first swing shorter than $1$ cm.

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Answers

a) $P(t) = -\frac{e^{-t}\left(\sin(\pi t) + \pi\cos(\pi t)\right)}{1 + \pi^2}$
b) $\int_0^1 v\,dt = \frac{\pi}{1 + \pi^2}(1 + e^{-1})$ m
c) $\int_1^2 v\,dt = -\frac{\pi}{1 + \pi^2}e^{-1}(1 + e^{-1})$ m; ratio $-e^{-1}$ .
d) Net $\frac{\pi}{1 + \pi^2}(1 - e^{-2})$ m; distance $\frac{\pi}{1 + \pi^2}(1 + e^{-1})^2$ m.
e) First swing about $0.40$ m; the swing during $[4, 5]$ is the first under $1$ cm (about $0.7$ cm).

a) Let $K = \int e^{-t}\sin(\pi t)\,dt$ . First trade: $u = \sin(\pi t)$ , $du = \pi\cos(\pi t)\,dt$ ; $dv = e^{-t}\,dt$ , $v = -e^{-t}$ : $K = -e^{-t}\sin(\pi t) + \pi\int e^{-t}\cos(\pi t)\,dt$ . Second trade, SAME type as $u$ : $u = \cos(\pi t)$ , $du = -\pi\sin(\pi t)\,dt$ , $v = -e^{-t}$ : $\int e^{-t}\cos(\pi t)\,dt = -e^{-t}\cos(\pi t) - \pi K$ . So $K = -e^{-t}\sin(\pi t) - \pi e^{-t}\cos(\pi t) - \pi^2 K$ , that is $(1 + \pi^2)K = -e^{-t}\left(\sin(\pi t) + \pi\cos(\pi t)\right)$ , and $P(t) = -\frac{e^{-t}\left(\sin(\pi t) + \pi\cos(\pi t)\right)}{1 + \pi^2}$ . Check: $P'(t) = \frac{e^{-t}(\sin \pi t + \pi\cos \pi t) - e^{-t}(\pi\cos \pi t - \pi^2 \sin \pi t)}{1 + \pi^2} = \frac{e^{-t}(1 + \pi^2)\sin \pi t}{1 + \pi^2} = v(t)$ . The number $1 + \pi^2$ is the $a^2 + b^2$ of exercise 4, with $a = -1$ and $b = \pi$ .

b) $P(1) = -\frac{e^{-1}(0 + \pi \cdot (-1))}{1 + \pi^2} = \frac{\pi e^{-1}}{1 + \pi^2}$ and $P(0) = -\frac{0 + \pi}{1 + \pi^2} = -\frac{\pi}{1 + \pi^2}$ . So $\int_0^1 v\,dt = P(1) - P(0) = \frac{\pi}{1 + \pi^2}(1 + e^{-1})$ metres. The sine vanishes at every integer, so only the cosine term survives at the bounds, and $\cos(\pi k) = (-1)^k$ does the rest: evaluating the bracket is quick once that is seen.

c) $P(2) = -\frac{e^{-2}(0 + \pi)}{1 + \pi^2}$ , so $\int_1^2 v\,dt = P(2) - P(1) = -\frac{\pi(e^{-2} + e^{-1})}{1 + \pi^2} = -\frac{\pi}{1 + \pi^2}e^{-1}(1 + e^{-1})$ metres. It is negative: the mass comes back, as the lobe below the axis shows. The ratio of the second displacement to the first is $-e^{-1}$ : the swing back is the first swing reversed and multiplied by $e^{-1} \approx 0.37$ .

d) Net displacement: $P(2) - P(0) = \frac{\pi - \pi e^{-2}}{1 + \pi^2} = \frac{\pi}{1 + \pi^2}(1 - e^{-2})$ metres, which is also the sum of b) and c), since $(1 + e^{-1}) - e^{-1}(1 + e^{-1}) = (1 + e^{-1})(1 - e^{-1}) = 1 - e^{-2}$ . The distance travelled adds the two swings in absolute value, because $v \ge 0$ on $[0, 1]$ and $v \le 0$ on $[1, 2]$ : $\frac{\pi}{1 + \pi^2}(1 + e^{-1}) + \frac{\pi}{1 + \pi^2}e^{-1}(1 + e^{-1}) = \frac{\pi}{1 + \pi^2}(1 + e^{-1})^2$ metres. Integrating $v$ straight over $[0, 2]$ gives the displacement, never the distance: the distance requires splitting at the zero of $v$ , $t = 1$ .

e) Since $\sin(\pi k) = 0$ and $\cos(\pi k) = (-1)^k$ , $P(k) = -\frac{\pi(-1)^k e^{-k}}{1 + \pi^2}$ , so $P(k + 1) - P(k) = \frac{\pi(-1)^k e^{-k}(1 + e^{-1})}{1 + \pi^2}$ , that is $(-1)^k e^{-k}$ times the first swing. Estimates: $\pi^2 \approx 9.87$ , so $\frac{\pi}{1 + \pi^2} \approx \frac{3.14}{10.87} \approx 0.29$ , and $1 + e^{-1} \approx 1.37$ : the first swing is about $0.40$ m. On the figure the first lobe rises to about $0.64$ over a base of $1$ s; a triangle would give $0.32$ , and the lobe is fuller than a triangle, so $0.40$ is plausible. The swings shrink by the factor $e$ each second: with $e^3 \approx 20$ and $e^4 \approx 55$ , the swing during $[3, 4]$ is about $\frac{0.40}{20} = 0.02$ m, still $2$ cm, and the one during $[4, 5]$ about $\frac{0.40}{55} \approx 0.007$ m. The first swing under $1$ cm is the fifth, during $[4, 5]$ .

Exercise 10: Final exam problem: x arctan x, the constant hidden in v, and an exact area

A typical final exam question, in four moves. The figure shows the region $A$ under $y = x\arctan x$ on $[0, 1]$ , and the dashed parabola $y = x^2$ .

Recall that $\arctan 1 = \frac{\pi}{4}$ and $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ .

a) Compute $\int x\arctan x\,dx$ with $u = \arctan x$ and $v = \frac{x^2}{2}$ .
b) Redo a) with $v = \frac{x^2 + 1}{2}$ . Why is it allowed, and what does it save?
c) Compute the area of $A$ exactly.
d) Prove that $\frac{\pi}{4}x \le \arctan x \le x$ on $[0, 1]$ , deduce bounds for the area of $A$ , and check c) against them.
e) Compute $\int_0^1 x^2\arctan x\,dx$ exactly. Does a clever constant in $v$ help this time?

Show the solution

Answers

a) $\int x\arctan x\,dx = \frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + C$
b) $v\,du = \frac{1}{2}\,dx$ : same answer in one line; any antiderivative of $x$ is a valid $v$ .
c) $A = \frac{\pi}{4} - \frac{1}{2}$
d) $\frac{\pi}{12} \le A \le \frac{1}{3}$ , and $\frac{\pi}{4} - \frac{1}{2} \approx 0.285$ lies between $0.262$ and $0.333$ .
e) $\int_0^1 x^2\arctan x\,dx = \frac{\pi}{12} - \frac{1}{6} + \frac{\ln 2}{6}$ ; no constant helps.

a) $u = \arctan x$ , $du = \frac{dx}{1 + x^2}$ ; $dv = x\,dx$ , $v = \frac{x^2}{2}$ . The inverse trigonometric factor is $u$ for the reason of exercise 1: its derivative is algebraic. Then $\int x\arctan x\,dx = \frac{x^2}{2}\arctan x - \frac{1}{2}\int \frac{x^2}{1 + x^2}\,dx$ . Add and subtract $1$ in the numerator: $\frac{x^2}{1 + x^2} = \frac{(1 + x^2) - 1}{1 + x^2} = 1 - \frac{1}{1 + x^2}$ . So the new integral is $x - \arctan x$ , and $\int x\arctan x\,dx = \frac{x^2}{2}\arctan x - \frac{x}{2} + \frac{1}{2}\arctan x + C = \frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + C$ .

b) Any function whose derivative is $x$ is a valid $v$ , since only $dv = x\,dx$ is prescribed (exercise 8, statement e). With $v = \frac{x^2 + 1}{2}$ : $\int x\arctan x\,dx = \frac{x^2 + 1}{2}\arctan x - \int \frac{x^2 + 1}{2} \cdot \frac{1}{1 + x^2}\,dx = \frac{x^2 + 1}{2}\arctan x - \int \frac{1}{2}\,dx = \frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + C$ . The constant $\frac{1}{2}$ was chosen so that $v$ contains the factor $1 + x^2$ of $du$ : the new integral collapses to a constant, and the algebra of a) disappears. The trade was already good; the constant made it free.

c) $x\arctan x \ge 0$ on $[0, 1]$ , so $A = \left[\frac{x^2 + 1}{2}\arctan x - \frac{x}{2}\right]_0^1 = \left(\frac{2}{2} \cdot \frac{\pi}{4} - \frac{1}{2}\right) - (0 - 0) = \frac{\pi}{4} - \frac{1}{2}$ . With $\pi \approx 3.14$ , $A \approx 0.285$ .

d) Upper bound: $g(x) = x - \arctan x$ has $g'(x) = 1 - \frac{1}{1 + x^2} = \frac{x^2}{1 + x^2} \ge 0$ and $g(0) = 0$ , so $g \ge 0$ on $[0, 1]$ , that is $\arctan x \le x$ . Lower bound: $(\arctan x)'' = -\frac{2x}{(1 + x^2)^2} \le 0$ on $[0, 1]$ , so $\arctan$ is concave there and lies above its chord from $(0, 0)$ to $(1, \frac{\pi}{4})$ , that is $\arctan x \ge \frac{\pi}{4}x$ . Multiplying by $x \ge 0$ : $\frac{\pi}{4}x^2 \le x\arctan x \le x^2$ , and integrating over $[0, 1]$ : $\frac{\pi}{12} \le A \le \frac{1}{3}$ . Numerically $0.262 \le 0.285 \le 0.333$ : c) passes. The figure shows the upper bound: the dashed parabola stays above the curve. A value like $\frac{\pi}{4} + \frac{1}{2}$ , from a sign slip, would be outside at once.

e) $u = \arctan x$ , $dv = x^2\,dx$ , $v = \frac{x^3}{3}$ : $\int_0^1 x^2\arctan x\,dx = \left[\frac{x^3}{3}\arctan x\right]_0^1 - \frac{1}{3}\int_0^1 \frac{x^3}{1 + x^2}\,dx$ . The bracket is $\frac{\pi}{12}$ . For the new integral, $x^3 = x(1 + x^2) - x$ , so $\frac{x^3}{1 + x^2} = x - \frac{x}{1 + x^2}$ , whose antiderivative is $\frac{x^2}{2} - \frac{1}{2}\ln(1 + x^2)$ (substitution $w = 1 + x^2$ ); on $[0, 1]$ it gives $\frac{1}{2} - \frac{\ln 2}{2}$ . Hence the value $\frac{\pi}{12} - \frac{1}{6} + \frac{\ln 2}{6}$ , about $0.262 - 0.167 + 0.115 = 0.21$ , below $A$ as it must be since $x^2 \le x$ on $[0, 1]$ . No constant helps here: $v = \frac{x^3 + K}{3}$ would need $x^3 + K$ to be a multiple of $1 + x^2$ , and no constant $K$ makes it so. The trick of b) works only when the antiderivative of $dv$ can be made to contain $1 + x^2$ .

Corrected exercises: integration by parts (MATH 141)

Course recap

Part A: the basics (/50)

Exercise 1: Choosing u and dv: the new integral is the judge

Exercise 2: One factor only: ln x, arctan x and arcsin x with dv = dx

Exercise 3: Repeated parts and the tabular method

Exercise 4: Cyclic parts: the integral that comes back

Exercise 5: Parts on a definite integral: the bracket is evaluated at both bounds

Part B: problems and reasoning (/50)

Exercise 6: A reduction formula: the trade written once for every n

Exercise 7: Parts when the function is unknown: moving the derivative

Exercise 8: Five statements to correct

Exercise 9: A damped oscillation: how far does the mass really travel?

Exercise 10: Final exam problem: x arctan x, the constant hidden in v, and an exact area

See also

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