Corrected exercises: the substitution rule (MATH 141)

a) The inside function is $x^3 + 1$, raised to the fifth power, and its derivative $3x^2$ is present up to the constant $3$. Let $u = x^3 + 1$, so $du = 3x^2\,dx$ and $x^2\,dx = \frac{1}{3}du$. Then $\int x^2(x^3 + 1)^5\,dx = \frac{1}{3}\int u^5\,du = \frac{1}{3}\cdot\frac{u^6}{6} + C = \frac{(x^3 + 1)^6}{18} + C$. Check: $\frac{d}{dx}\frac{(x^3+1)^6}{18} = \frac{6(x^3 + 1)^5 \cdot 3x^2}{18} = x^2(x^3 + 1)^5$. The constant $\frac{1}{3}$ is legitimate because it is a NUMBER: it can be moved through the integral sign. What cannot be moved is a factor containing $x$, and this is exactly why the $x^2$ outside the bracket is the whole reason the substitution works. Expanding $(x^3 + 1)^5$ would also succeed, but costs six terms and ten minutes.

b) The inside function is $\sqrt{x}$, and its derivative $\frac{1}{2\sqrt{x}}$ sits in the integrand up to the constant $\frac{1}{2}$. Let $u = \sqrt{x}$, so $du = \frac{dx}{2\sqrt{x}}$ and $\frac{dx}{\sqrt{x}} = 2\,du$. Then $\int \frac{\cos\sqrt{x}}{\sqrt{x}}\,dx = 2\int \cos u\,du = 2\sin u + C = 2\sin\sqrt{x} + C$. Check: $\frac{d}{dx}\,2\sin\sqrt{x} = 2\cos\sqrt{x}\cdot\frac{1}{2\sqrt{x}}$, the integrand. The reflex $u = \cos\sqrt{x}$ goes nowhere: its derivative $-\sin\sqrt{x}\cdot\frac{1}{2\sqrt x}$ is not in the integrand. The rule is to choose the INSIDE of the composition, not the outside function.

c) The inside function is $\ln x$, and its derivative $\frac{1}{x}$ is the other factor. Let $u = \ln x$, $du = \frac{dx}{x}$: $\int \frac{(\ln x)^3}{x}\,dx = \int u^3\,du = \frac{u^4}{4} + C = \frac{(\ln x)^4}{4} + C$. Check: $\frac{d}{dx}\frac{(\ln x)^4}{4} = (\ln x)^3\cdot\frac{1}{x}$. A $\frac{1}{x}$ standing next to a function of $\ln x$ is the most reliable signal of the chapter: it is $du$ already written for you.

d) The inside function is $9 - x^2$, whose derivative $-2x$ is present up to the constant $-2$. Let $u = 9 - x^2$, $du = -2x\,dx$, so $x\,dx = -\frac{1}{2}du$ and $\int \frac{x}{\sqrt{9 - x^2}}\,dx = -\frac{1}{2}\int u^{-1/2}\,du = -\frac{1}{2}\cdot 2u^{1/2} + C = -\sqrt{9 - x^2} + C$. Check: $\frac{d}{dx}\left(-\sqrt{9 - x^2}\right) = -\frac{-2x}{2\sqrt{9 - x^2}} = \frac{x}{\sqrt{9 - x^2}}$. Without the factor $x$ on top, $u = 9 - x^2$ fails: $du$ would need an $x$ that is not there, and dividing by it is forbidden. The integral $\int \frac{dx}{\sqrt{9 - x^2}}$ is instead a table integral in disguise. Factor $9$: $\sqrt{9 - x^2} = 3\sqrt{1 - (x/3)^2}$, then $w = \frac{x}{3}$, $dx = 3\,dw$, gives $\int \frac{3\,dw}{3\sqrt{1 - w^2}} = \arcsin w + C = \arcsin\frac{x}{3} + C$. One factor $x$ changes the answer from a square root to an inverse sine.

e) With $u = \sin x$, $du = \cos x\,dx$: $\int \sin x\cos x\,dx = \int u\,du = \frac{\sin^2 x}{2} + C_1$. With $u = \cos x$, $du = -\sin x\,dx$: $\int \sin x\cos x\,dx = -\int u\,du = -\frac{\cos^2 x}{2} + C_2$. Both are right, since both differentiate back to $\sin x\cos x$. They are not contradictory, because $\frac{\sin^2 x}{2} - \left(-\frac{\cos^2 x}{2}\right) = \frac{\sin^2 x + \cos^2 x}{2} = \frac{1}{2}$, a constant. Two antiderivatives of the same function on an interval differ by a constant, and the $+C$ absorbs it. A third route, $\sin x\cos x = \frac{1}{2}\sin 2x$ with $w = 2x$, gives $-\frac{\cos 2x}{4} + C_3$, again off by a constant. On an exam, any of the three earns full marks; what loses the mark is dropping the $+ C$, which is the only thing that makes the three answers the same answer.

a) Let $u = 1 + x^2$, $du = 2x\,dx$, so $x\,dx = \frac{1}{2}du$. Route 1: $\int x\sqrt{1 + x^2}\,dx = \frac{1}{2}\int u^{1/2}\,du = \frac{1}{3}u^{3/2} + C = \frac{1}{3}(1 + x^2)^{3/2} + C$, then $\left[\frac{1}{3}(1 + x^2)^{3/2}\right]_0^{\sqrt 3} = \frac{1}{3}\left(4^{3/2} - 1^{3/2}\right) = \frac{1}{3}(8 - 1) = \frac{7}{3}$. Route 2: when $x = 0$, $u = 1$; when $x = \sqrt 3$, $u = 4$. So $\int_0^{\sqrt 3} x\sqrt{1 + x^2}\,dx = \frac{1}{2}\int_1^4 u^{1/2}\,du = \frac{1}{3}\left[u^{3/2}\right]_1^4 = \frac{7}{3}$. The same number, as it must be. Route 2 is shorter and is the one to use on an exam, provided the bound line (when $x = \ldots$, $u = \ldots$) is WRITTEN: it is the line the marker looks for.

b) Let $u = \ln x$, $du = \frac{dx}{x}$. When $x = 1$, $u = \ln 1 = 0$; when $x = e^2$, $u = 2$. Then $\int_1^{e^2}\frac{\sqrt{\ln x}}{x}\,dx = \int_0^2 u^{1/2}\,du = \frac{2}{3}\left[u^{3/2}\right]_0^2 = \frac{2}{3}\cdot 2\sqrt 2 = \frac{4\sqrt 2}{3}$. The bounds $1$ and $e^2$ were chosen for the logarithm: without a calculator, $\ln(e^2) = 2$ is the kind of bound an exam uses, and it only becomes simple AFTER the substitution. Order of magnitude: $\sqrt 2 \approx 1.41$, so the answer is about $1.89$.

c) Let $u = \cos x$, $du = -\sin x\,dx$, so $\sin x\,dx = -du$. When $x = 0$, $u = \cos 0 = 1$; when $x = \frac{\pi}{2}$, $u = 0$. Then $\int_0^{\pi/2}\sin x\,e^{\cos x}\,dx = \int_1^0 e^u(-du) = -\int_1^0 e^u\,du = \int_0^1 e^u\,du = e - 1$. The new lower bound $1$ is LARGER than the new upper bound $0$, and that is fine: keep the bounds in the order the substitution gives them. The minus sign of $du$ and the reversed order cancel, since $-\int_1^0 = \int_0^1$. The classic slip is to use one of them and not the other: reorder the bounds to $\int_0^1$ AND keep the minus, and the answer becomes $1 - e$. The figure refutes it at once: the integrand is positive on $\left(0, \frac{\pi}{2}\right)$, the shaded area is positive, so the integral is positive. With $e \approx 2.72$, $e - 1 \approx 1.72$, plausible for a region about $1.57$ wide and at most about $1.5$ high.

d) The inside function is $1 + \sqrt{x}$, and its derivative $\frac{1}{2\sqrt x}$ is present up to the constant $\frac{1}{2}$. Let $u = 1 + \sqrt{x}$, $du = \frac{dx}{2\sqrt x}$, so $\frac{dx}{\sqrt x} = 2\,du$. When $x = 1$, $u = 2$; when $x = 4$, $u = 3$. Then $\int_1^4\frac{dx}{\sqrt x\,(1 + \sqrt x)^2} = 2\int_2^3 u^{-2}\,du = 2\left[-\frac{1}{u}\right]_2^3 = 2\left(-\frac{1}{3} + \frac{1}{2}\right) = \frac{1}{3}$. The new bounds are $1 + \sqrt 1$ and $1 + \sqrt 4$, not $\sqrt 1$ and $\sqrt 4$: a bound is always the value of the WHOLE $u$, constant included.

e) She has substituted in the integrand and the differential but not in the bounds: $\frac{1}{2}\int_0^{\sqrt 3}u^{1/2}\,du = \frac{1}{3}\left[u^{3/2}\right]_0^{\sqrt 3} = \frac{1}{3}(\sqrt 3)^{3/2} = \frac{3^{3/4}}{3}$, about $0.76$, instead of $\frac{7}{3} \approx 2.33$. The numbers $0$ and $\sqrt 3$ are values of $x$; the variable of her new integral is $u$, which runs from $1$ to $4$. Her integral still contains an $x$, hidden in the bounds, and that is exactly the unfinished substitution. The symmetric mixture is just as wrong: change the bounds to $1$ and $4$, then substitute back to $x$ and evaluate $\frac{1}{3}\left[(1 + x^2)^{3/2}\right]_1^4 = \frac{1}{3}\left(17^{3/2} - 8\right)$, which is about $20.7$. Choose a route and follow it to the end: bounds in $x$ with an antiderivative in $x$, or bounds in $u$ with an antiderivative in $u$.

a) Let $u = x - 1$, $du = dx$. The factor $\sqrt{x - 1}$ becomes $\sqrt u$, but a factor $x$ remains. From $u = x - 1$: $x = u + 1$. So $\int x\sqrt{x - 1}\,dx = \int (u + 1)u^{1/2}\,du = \int \left(u^{3/2} + u^{1/2}\right)du = \frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2} + C = \frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C$. Check: the derivative is $(x - 1)^{3/2} + (x - 1)^{1/2} = (x - 1)^{1/2}\left[(x - 1) + 1\right] = x\sqrt{x - 1}$. The substitution does not simplify the root, it moves the difficulty to a place where the power rule applies: $(u + 1)\sqrt u$ expands, $x\sqrt{x - 1}$ does not.

b) Let $u = x + 1$, $du = dx$, $x = u - 1$. When $x = 0$, $u = 1$; when $x = 3$, $u = 4$. Then $\int_0^3\frac{x}{\sqrt{x + 1}}\,dx = \int_1^4\frac{u - 1}{\sqrt u}\,du = \int_1^4\left(u^{1/2} - u^{-1/2}\right)du = \left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]_1^4 = \left(\frac{16}{3} - 4\right) - \left(\frac{2}{3} - 2\right) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$. The division $\frac{u - 1}{\sqrt u}$ is term by term, $\frac{u}{\sqrt u} = u^{1/2}$ and $\frac{1}{\sqrt u} = u^{-1/2}$. Sanity check: the integrand grows from $0$ to $\frac{3}{2}$ on an interval of length $3$, so the integral lies between $0$ and $4.5$; $\frac{8}{3} \approx 2.67$ fits.

c) The inside function is $x^2 + 1$, with derivative $2x$. Write $x^3\,dx = x^2\cdot x\,dx$: ONE factor $x$ goes with $dx$ to form $du$, the other two form $x^2$, which is rewritten. Let $u = x^2 + 1$, $du = 2x\,dx$, $x\,dx = \frac{1}{2}du$, $x^2 = u - 1$. Then $\int x^3\sqrt{x^2 + 1}\,dx = \frac{1}{2}\int (u - 1)u^{1/2}\,du = \frac{1}{2}\left(\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right) + C = \frac{1}{5}(x^2 + 1)^{5/2} - \frac{1}{3}(x^2 + 1)^{3/2} + C$. Check: the derivative is $x(x^2 + 1)^{3/2} - x(x^2 + 1)^{1/2} = x(x^2 + 1)^{1/2}\left[(x^2 + 1) - 1\right] = x^3\sqrt{x^2 + 1}$. The split works because the leftover power is EVEN: $x^2$ is a function of $x^2 + 1$. With $x^2\sqrt{x^2 + 1}$ the leftover would be a single $x = \sqrt{u - 1}$, and that integral belongs to the trigonometric substitution chapter.

d) Differentiate her answer with the product rule: $\frac{d}{dx}\left[\frac{2}{3}x(x - 1)^{3/2}\right] = \frac{2}{3}(x - 1)^{3/2} + \frac{2}{3}x\cdot\frac{3}{2}(x - 1)^{1/2} = \frac{2}{3}(x - 1)^{3/2} + x\sqrt{x - 1}$. The second term is the integrand; the first, $\frac{2}{3}(x - 1)^{3/2}$, is extra, and it is precisely the derivative of the $x$ she froze. At $x = 2$ her derivative is $\frac{2}{3} + 2 = \frac{8}{3}$ where the integrand is $2$. Only a CONSTANT passes through the integral sign; $x$ varies, so $\int x\,h(x)\,dx \ne x\int h(x)\,dx$. The correct answer is that of part a).

e) Let $u = 1 + \sqrt{x}$. Here $du = \frac{dx}{2\sqrt x}$ does not appear in the integrand, so solve for $x$ instead: $\sqrt{x} = u - 1$, $x = (u - 1)^2$ and $dx = 2(u - 1)\,du$. When $x = 0$, $u = 1$; when $x = 4$, $u = 3$. Then $\int_0^4\frac{dx}{1 + \sqrt x} = \int_1^3\frac{2(u - 1)}{u}\,du = \int_1^3\left(2 - \frac{2}{u}\right)du = \left[2u - 2\ln u\right]_1^3 = (6 - 2\ln 3) - (2 - 0) = 4 - 2\ln 3$. With $\ln 3 \approx 1.10$ that is about $1.80$, and the integrand falls from $1$ to $\frac{1}{3}$ over a length $4$, so the value must lie between $\frac{4}{3}$ and $4$: it does. Rewriting $dx$ through $u$ is the same gesture as rewriting a stray $x$: whatever is left in $x$ is expressed in $u$.

a) Write $\tan x = \frac{\sin x}{\cos x}$. The derivative of the denominator is $-\sin x$, the numerator up to the constant $-1$. Let $u = \cos x$, $du = -\sin x\,dx$: $\int \tan x\,dx = -\int \frac{du}{u} = -\ln|\cos x| + C$. Since $-\ln|\cos x| = \ln\frac{1}{|\cos x|}$, the same answer reads $\ln|\sec x| + C$, the form of Stewart's table. For the cotangent, $\cot x = \frac{\cos x}{\sin x}$ and $u = \sin x$, $du = \cos x\,dx$ give $\int \cot x\,dx = \ln|\sin x| + C$, with no minus sign. The two signs are opposite because the derivatives of $\cos$ and $\sin$ have opposite signs; forgetting the minus of $d(\cos x)$ is the most common error on this integral, and part b) shows how to catch it.

b) $\int_0^{\pi/3}\tan x\,dx = \left[-\ln|\cos x|\right]_0^{\pi/3} = -\ln\frac{1}{2} + \ln 1 = \ln 2$. On $\left(0, \frac{\pi}{3}\right)$ the tangent is positive, the shaded region of the figure lies above the axis, so the integral must be positive: $\ln 2 \approx 0.69$ is, and the region, about $1.05$ wide and at most $\sqrt 3 \approx 1.73$ high but mostly much lower, is consistent with it. The wrong sign convention, $\ln|\cos x|$, gives $-\ln 2$, negative for a positive integrand: reject it on sight. Then $\int_{\pi/6}^{\pi/2}\cot x\,dx = \left[\ln|\sin x|\right]_{\pi/6}^{\pi/2} = \ln 1 - \ln\frac{1}{2} = \ln 2$. The two answers agree, and they should: the substitution $x \mapsto \frac{\pi}{2} - x$ exchanges $\tan$ and $\cot$ and maps $\left[0, \frac{\pi}{3}\right]$ onto $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$.

c) The derivative of $x^2 + 3x + 5$ is exactly $2x + 3$. With $u = x^2 + 3x + 5$, $du = (2x + 3)\,dx$ and $\int \frac{2x + 3}{x^2 + 3x + 5}\,dx = \ln|x^2 + 3x + 5| + C$. The discriminant of the denominator is $9 - 20 = -11 < 0$ and its leading coefficient is positive, so $x^2 + 3x + 5 > 0$ for every $x$ (it equals $\left(x + \frac{3}{2}\right)^2 + \frac{11}{4}$), and the answer may be written $\ln(x^2 + 3x + 5) + C$. Likewise $u = e^x + 3$, $du = e^x\,dx$ give $\int \frac{e^x}{e^x + 3}\,dx = \ln(e^x + 3) + C$, and $e^x + 3 > 3$ needs no absolute value. Dropping the bars is allowed only after a sentence that proves the argument is positive; without that sentence, keep them.

d) Let $u = x^2 - 4$, $du = 2x\,dx$. When $x = -1$, $u = -3$; when $x = 0$, $u = -4$. Then $\int_{-1}^0\frac{2x}{x^2 - 4}\,dx = \int_{-3}^{-4}\frac{du}{u} = \left[\ln|u|\right]_{-3}^{-4} = \ln 4 - \ln 3 = \ln\frac{4}{3}$. The student used $\ln u$ where the antiderivative of $\frac{1}{u}$ is $\ln|u|$: on the interval $[-4, -3]$, where $u$ is negative, $\ln|u| = \ln(-u)$ is perfectly defined and its derivative is $\frac{-1}{-u} = \frac{1}{u}$. The integrand is continuous on $[-1, 0]$ (the denominator vanishes only at $\pm 2$), so the integral certainly exists. Sign check: on $(-1, 0)$ the numerator $2x$ and the denominator $x^2 - 4$ are both negative, the integrand is positive, and $\ln\frac{4}{3} > 0$.

e) The numerator is not visibly the derivative of the denominator, until the fraction is read as $\frac{1/x}{\ln x}$: the derivative of $\ln x$ is $\frac{1}{x}$. Let $u = \ln x$, $du = \frac{dx}{x}$; when $x = e$, $u = 1$; when $x = e^3$, $u = 3$. Then $\int_e^{e^3}\frac{dx}{x\ln x} = \int_1^3\frac{du}{u} = \ln 3 - \ln 1 = \ln 3$. As for $\int \frac{dx}{1 + x^2}$: the derivative of $\ln(1 + x^2)$ is $\frac{2x}{1 + x^2}$, not $\frac{1}{1 + x^2}$. The rule $\ln|g|$ needs $g'$ ON TOP, and here the numerator $1$ is not a constant multiple of $2x$. No substitution is needed at all: $\int \frac{dx}{1 + x^2} = \arctan x + C$ is in the table. The same denominator gives a logarithm with $x$ on top and an inverse tangent with $1$ on top, and exercise 6 turns that difference into a method.