Integration: substitution and parts

Worked solutions

Problem 1

1. $u = x^2$, $du = 2x\,dx$.$e^{x^2}$ is $e^u$ with inner function $x^2$, and the integrand also contains $x$, which is half of that inner function's derivative $2x$.
2. $x\,dx = \tfrac12\,du$.Divide both sides of step 1 by 2, so that the $x\,dx$ in the integral can be replaced as a whole.
3. $\displaystyle\int xe^{x^2}\,dx = \int e^u\cdot\tfrac12\,du = \tfrac12e^u + C$.Substitute steps 1 and 2; $(e^u)' = e^u$, and the constant $\tfrac12$ comes out of the integral.
4. $\tfrac12e^{x^2} + C$Put $u = x^2$ back. Check: by the chain rule, $\tfrac{d}{dx}\tfrac12e^{x^2} = \tfrac12e^{x^2}\cdot 2x = xe^{x^2}$.

Problem 2

1. $u = 1 + x^2$, $du = 2x\,dx$.The numerator $2x\,dx$ is exactly the derivative of the denominator, times $dx$.
2. $\displaystyle\int \frac{2x}{1+x^2}\,dx = \int\frac{du}{u} = \ln|u| + C$.The denominator becomes $u$ and the numerator with $dx$ becomes $du$.
3. $\ln(1+x^2) + C$Put $u = 1 + x^2$ back; $1 + x^2 \geq 1 > 0$ for every $x$, so the absolute value can be dropped and the answer holds on the whole real line.

Problem 3

1. $u = x$, $dv = e^x\,dx$; $du = dx$, $v = e^x$.Differentiating $x$ makes it simpler, the constant 1, and $e^x$ is its own antiderivative, so the new integral $\int v\,du$ has no factor of $x$ left.
2. $\displaystyle\int xe^x\,dx = xe^x - \int e^x\,dx$.Parts: $uv - \int v\,du$.
3. $xe^x - e^x + C$.$\int e^x\,dx = e^x + C$.
4. $(x - 1)e^x + C$Factor out $e^x$. Check: by the product rule, $\big((x-1)e^x\big)' = e^x + (x-1)e^x = xe^x$.

Problem 4

1. $u = \ln x$, $dv = dx$; $du = \dfrac{dx}{x}$, $v = x$.No antiderivative of $\ln x$ is known yet, but its derivative, $1/x$, is simple; the other factor is then 1, and $dx$ integrates to $x$. $\ln x$ needs $x > 0$.
2. $\displaystyle\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx$.Parts: $uv - \int v\,du$.
3. $x\ln x - \int 1\,dx = x\ln x - x + C$.$x \cdot \tfrac1x = 1$ for $x > 0$, and $\int 1\,dx = x + C$.
4. $x\ln x - x + C$Check: $(x\ln x - x)' = \ln x + x\cdot\tfrac1x - 1 = \ln x$.

Problem 5

1. $u = x$, $dv = \cos x\,dx$; $du = dx$, $v = \sin x$.Differentiating $x$ removes it, and $\cos x$ integrates to $\sin x$ without getting harder.
2. $\displaystyle\int x\cos x\,dx = x\sin x - \int \sin x\,dx$.Parts: $uv - \int v\,du$.
3. $\displaystyle\int \sin x\,dx = -\cos x + C$.$(-\cos x)' = \sin x$.
4. $x\sin x + \cos x + C$Subtracting $-\cos x$ adds $\cos x$. Check: $(x\sin x + \cos x)' = \sin x + x\cos x - \sin x = x\cos x$.

Problem 6

1. Call the integral $I = \displaystyle\int e^x\sin x\,dx$.Neither factor gets simpler when differentiated, so parts will not finish the job; the plan is to apply parts twice until $I$ reappears, then solve for it.
2. $u = \sin x$, $dv = e^x\,dx$; $du = \cos x\,dx$, $v = e^x$.$e^x$ integrates to itself, so $v$ stays as simple as $dv$.
3. $I = e^x\sin x - \displaystyle\int e^x\cos x\,dx$.Parts: $uv - \int v\,du$.
4. For $\displaystyle\int e^x\cos x\,dx$: $u = \cos x$, $dv = e^x\,dx$; $du = -\sin x\,dx$, $v = e^x$.Keep the same choice, trig as $u$ and the exponential as $dv$. Swapping it would integrate the $\cos x$ back to $\sin x$ and undo step 3, giving only $I = I$.
5. $\displaystyle\int e^x\cos x\,dx = e^x\cos x - \int e^x(-\sin x)\,dx = e^x\cos x + I$.Parts again; the minus from $du$ turns the new integral into $+I$.
6. $I = e^x\sin x - e^x\cos x - I$.Substitute step 5 into step 3. The original integral now appears on both sides.
7. $2I = e^x(\sin x - \cos x)$.Add $I$ to both sides and factor out $e^x$. The constants of integration from the two applications of parts are gathered into one $C$ in the final line.
8. $\tfrac12e^x(\sin x - \cos x) + C$Divide by 2. Check: $\big(\tfrac12e^x(\sin x - \cos x)\big)' = \tfrac12e^x(\sin x - \cos x) + \tfrac12e^x(\cos x + \sin x) = e^x\sin x$.

Problem 7

1. $y = \arctan x$ means $\tan y = x$ with $-\tfrac\pi2 < y < \tfrac\pi2$.$\arctan$ is the inverse of $\tan$ on $(-\tfrac\pi2, \tfrac\pi2)$, where $\cos y > 0$ and $\tan$ is increasing, so each $x$ has exactly one $y$.
2. $\dfrac{1}{\cos^2 y}\,y' = 1$.Differentiate $\tan y = x$ with respect to $x$: the chain rule gives the derivative of $\tan$ at $y$, which is $1/\cos^2 y$ (quotient rule on $\sin y/\cos y$), times $y'$.
3. $y' = \cos^2 y = \dfrac{1}{1 + \tan^2 y} = \dfrac{1}{1+x^2}$.$1 + \tan^2 y = \dfrac{\cos^2 y + \sin^2 y}{\cos^2 y} = \dfrac{1}{\cos^2 y}$, and $\tan y = x$.
4. $\displaystyle\int_0^1 \frac{dx}{1+x^2} = \big[\arctan x\big]_0^1 = \arctan 1 - \arctan 0$.Step 3 makes $\arctan x$ an antiderivative of the integrand, and a definite integral is the antiderivative's value at the upper limit minus its value at the lower one.
5. $\arctan 1 = \tfrac\pi4$ and $\arctan 0 = 0$.$\tan\tfrac\pi4 = 1$ and $\tan 0 = 0$, and both angles lie in $(-\tfrac\pi2, \tfrac\pi2)$.
6. $\dfrac\pi4$$\tfrac\pi4 - 0$. The integrand lies between $\tfrac12$ and 1 on $[0, 1]$, and $\tfrac\pi4 \approx 0.785$ is between them.

Problem 8

1. $\displaystyle\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx$, on an interval where $\cos x \neq 0$.$\tan x$ is defined only where $\cos x \neq 0$; written as a quotient, the numerator is, up to sign, the derivative of the denominator.
2. $u = \cos x$, $du = -\sin x\,dx$, so $\sin x\,dx = -du$.$(\cos x)' = -\sin x$; the minus sign has to be carried into the integral.
3. $\displaystyle\int \frac{\sin x}{\cos x}\,dx = \int\frac{-du}{u} = -\ln|u| + C$.$\int du/u = \ln|u| + C$ for $u \neq 0$; the absolute value is needed because $\cos x$ is negative on half of the intervals where $\tan x$ is defined.
4. $-\ln|\cos x| + C$Put $u = \cos x$ back. Check: where $\cos x > 0$, the chain rule gives $\tfrac{d}{dx}\ln\cos x = \dfrac{-\sin x}{\cos x} = -\tan x$; where $\cos x < 0$, it gives $\tfrac{d}{dx}\ln(-\cos x) = \dfrac{\sin x}{-\cos x} = -\tan x$. Either way the minus sign in front makes the derivative $\tan x$.

Problem 9

1. $\displaystyle\int_0^\infty xe^{-x}\,dx = \lim_{b\to\infty}\int_0^b xe^{-x}\,dx$.The upper limit is not a number, so the integral is defined as the limit of integrals over $[0, b]$.
2. $u = x$, $dv = e^{-x}\,dx$; $du = dx$, $v = -e^{-x}$.Differentiating $x$ removes it; $(-e^{-x})' = e^{-x}$ by the chain rule.
3. $\displaystyle\int_0^b xe^{-x}\,dx = \big[-xe^{-x}\big]_0^b + \int_0^b e^{-x}\,dx$.Parts on $[0, b]$: $\big[uv\big]_0^b - \int_0^b v\,du$, and $-v = e^{-x}$.
4. $\big[-xe^{-x}\big]_0^b = -be^{-b} - 0$ and $\displaystyle\int_0^b e^{-x}\,dx = \big[-e^{-x}\big]_0^b = 1 - e^{-b}$.Evaluate each bracket at $b$ and at 0; at $x = 0$ the first bracket is $0\cdot e^0 = 0$.
5. $\displaystyle\int_0^b xe^{-x}\,dx = 1 - e^{-b} - be^{-b}$.Add the two parts of step 4.
6. $0 < be^{-b} = \dfrac{b}{e^b} < \dfrac{b}{b^2/2} = \dfrac2b$ for $b > 0$, so $be^{-b} \to 0$.$e^b = 1 + b + \tfrac{b^2}{2} + \cdots > \tfrac{b^2}{2}$ for $b > 0$, since every term of the series is positive; the bound $2/b$ goes to 0.
7. $1$Take $b \to \infty$ in step 5: $e^{-b} \to 0$ and, by step 6, $be^{-b} \to 0$. This is the mean of the exponential distribution with rate 1.

Problem 10

1. $\sigma(x) = \dfrac{e^x}{1+e^x}$.Multiply top and bottom by $e^x$, which is never 0, so that the numerator becomes the derivative of the denominator.
2. $u = 1 + e^x$, $du = e^x\,dx$.The denominator's derivative, $e^x$, is the numerator.
3. $\displaystyle\int \frac{e^x}{1+e^x}\,dx = \int\frac{du}{u} = \ln|u| + C$.Substitute step 2.
4. $\displaystyle\int \sigma(x)\,dx = \ln(1+e^x) + C$.Put $u = 1 + e^x$ back; $1 + e^x > 1$, so the absolute value can be dropped. This is softplus, whose derivative is $\sigma$ (Differentiation rules, Problem 4).
5. $\displaystyle\int_{-\infty}^0\sigma(x)\,dx = \lim_{a\to-\infty}\int_a^0\sigma(x)\,dx = \lim_{a\to-\infty}\big(\ln 2 - \ln(1+e^a)\big)$.The lower limit is not a number, so the integral is a limit; step 4 evaluated at $0$ gives $\ln 2$, since $1 + e^0 = 2$, and at $a$ gives $\ln(1+e^a)$.
6. $\ln(1+e^a) \to \ln 1 = 0$ as $a \to -\infty$.$e^a \to 0$, and $\ln$ is continuous at 1.
7. $\ln(1+e^x) + C$; and $\displaystyle\int_{-\infty}^0\sigma = \ln 2 - \lim_{a\to-\infty}\ln(1+e^a) = \ln 2$Steps 4 to 6. The area is finite although $\sigma$ never reaches 0, because $\sigma(x) < e^x$ decays exponentially as $x \to -\infty$.