Corrected exercises: integration by parts (MATH 141)

$u = x$, $du = dx$; $dv = e^{2x}\,dx$, $v = \frac{1}{2}e^{2x}$ (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.

$u = e^{2x}$, $du = 2e^{2x}\,dx$; $dv = x\,dx$, $v = \frac{x^2}{2}$. 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.

$u = \ln x$, $du = \frac{dx}{x}$; $dv = x^3\,dx$, $v = \frac{x^4}{4}$. 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.

$u = x$, $du = dx$; $dv = 2^x\,dx$, $v = \frac{2^x}{\ln 2}$, 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.

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.

$u = \ln x$, $du = \frac{dx}{x}$; $dv = dx$, $v = x$. 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: $\frac{d}{dx}(x \ln x - x) = \ln x + 1 - 1 = \ln x$.

$u = \arctan x$, $du = \frac{dx}{1 + x^2}$; $dv = dx$, $v = x$. 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 $dv = dx$.

$u = \arcsin x$, $du = \frac{dx}{\sqrt{1 - x^2}}$; $dv = dx$, $v = x$. 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.

$\ln x \ge 0$ 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.

$\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$: 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.

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.

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.

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.

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.

$\frac{d}{dx}\left[-e^{-x}(x^2 - 2x + 2)\right] = e^{-x}(x^2 - 2x + 2) - e^{-x}(2x - 2) = e^{-x}(x^2 - 4x + 4) = e^{-x}(x - 2)^2$, 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.

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: $\frac{1}{2}\left[e^x(\sin x - \cos x) + e^x(\cos x + \sin x)\right] = e^x \sin x$.

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.

$e^x > 0$ 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.

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.

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.