4.3 Substitution

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In this section we learn to reverse the chain rule by making a substitution.

U-Substitution

We make a u-substitution to fill in the gaps in the following equation, which reverses the chain rule:

$\int f(g(x))g'(x) \ dx = F(g(x)) + C.$

If we let $u = g(x)$ then $du = g'(x) \ dx$ and the integral can be rewritten as $\int f(g(x))g'(x) \ dx = \int f(u) \ du$ which equals $F(u) + C$ and back substituting gives $F(g(x)) + C.$ Now we can see that $\int f(g(x))g'(x) \ dx = F(g(x)) + C.$

Compute $\int 2x\cos (x^2 + 1) \ dx.$

Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

and the integral can be written as $\int 2x\cos (x^2 + 1) \ dx = \int \cos (x^2 + 1) \ 2x\ dx = \int \cos (u) \ du.$ The last integral can be computed as $\int \cos (u) \ du = \sin (u) + C$ and by back substituting, we have $\sin (u) + C = \sin (x^2 + 1) + C.$ Thus, using u-substitution we can conclude that $\int 2x\cos (x^2 + 1) \ dx = \sin (x^2 + 1) + C.$

Let $u = x^3 + 2$

Compute $du$

$\int 3x^2\cos (x^3 + 2) \ dx = \answer {\sin (x^3 + 2)} +C.$

Let $u = x^5 -7$

Compute $du$

$\int 5x^4\sin (x^5 -7) \ dx = \answer {-\cos (x^5 -7)} +C.$

Compute $\int 2x(x^2 + 1)^4 \ dx.$ Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

and the integral can be written as $\int 2x(x^2 + 1)^4 \ dx = \int (x^2 + 1)^4 \ 2x \ dx = \int u^4 \ du.$ The last integral can be computed as $\int u^4 \ du = \tfrac {u^5}{5} + C$ and by back substituting, we have $\tfrac {u^5}{5} + C = \tfrac 15(x^2 + 1)^5 + C .$ Thus, using u-substitution we can conclude that $\int 2x(x^2 + 1)^4 \ dx = \tfrac 15(x^2 + 1)^5 + C .$

Let $u = x^3 + 2$

Compute $du$

$\int 3x^2(x^3 + 2)^6 \ dx = \answer {(x^3 + 2)^7/7} +C.$

Let $u = x^4 -1$

Compute $du$

$\int 4x^3(x^4 -1)^8 \ dx = \answer {(x^4 -1)^9/9} +C.$

Compute $\int 2xe^{(x^2 + 1)} \ dx.$ Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

$\int 2xe^{(x^2 + 1)} \ dx = \int e^{(x^2 + 1)} \ 2x\ dx = \int e^u \ du.$ The last integral can be computed as $\int e^u \ du = e^u + C$ and by back substituting, we have $e^u + C = e^{(x^2 + 1)} + C.$ Thus, using u-substitution we can conclude that $\int 2xe^{(x^2 + 1)} \ dx = e^{(x^2 + 1)} + C.$

Let $u = x^3 + 2$

Compute $du$

$\int 3x^2e^{x^3 + 2} \ dx = \answer {e^{x^3 + 2}} +C.$

Let $u = -x^2$

Compute $du$

$\int -2xe^{-x^2} \ dx = \answer {e^{-x^2}} +C.$

Compute $\int 2x\sqrt {x^2 + 1} \ dx.$ Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

$\int 2x\sqrt {x^2 + 1} \ dx = \int \sqrt {x^2 + 1} \cdot 2x\ dx = \int \sqrt {u} \ du.$ The last integral can be computed as $\int \sqrt u \ du = \int u^{1/2} \ du = \tfrac {u^{3/2}}{3/2} + C = \tfrac 23 u^{3/2} + C$ and by back substituting, we have $\tfrac 23 u^{3/2} + C = \frac 23 (x^2 + 1)^{3/2} + C.$ Thus, using u-substitution we can conclude that $\int 2x\sqrt {x^2 + 1} \ dx = \tfrac 23 (x^2 + 1)^{3/2} + C.$

Let $u = x^4 + 2$

Compute $du$

$\int 4x^3 \sqrt {x^4 +2} \ dx = \answer {2/3(x^4 +2)^{3/2}} +C.$

Compute $\int \frac {2x}{x^2 + 1} \ dx.$ Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

$\int \frac {2x}{x^2 + 1} \ dx = \int \frac {1}{x^2 + 1} \cdot 2x\ dx = \int \frac {1}{u} \ du.$ The last integral can be computed as $\int \frac {1}{u} \ du = \ln |u| + C$ and by back substituting, we have $\ln |u| + C = \ln |x^2 + 1| + C=\ln (x^2 + 1) + C.$ Thus, using u-substitution we can conclude that $\int \frac {2x}{x^2 + 1} \ dx = \ln (x^2 + 1) + C.$

Let $u = x^4 + 2$

Compute $du$

$\int \frac {4x^3}{x^4 +2} \ dx = \answer {\ln |x^4 +2|} +C.$

Compute $\int e^x\sin (e^x) \ dx.$ We let $u = e^x$ . We have, $u = e^x$ $du = e^x \ dx$

and the integral can be written as $\int e^x\sin (e^x) \ dx = \int \sin (e^x) \cdot e^x \ dx = \int \sin (u) \ du.$ The last integral can be computed as $\int \sin (u) \ du = -\cos (u) + C$ and by back substituting, we have $-\cos (u) + C = -\cos (e^x) + C.$ Thus, using u-substitution we can conclude that $\int e^x\sin (e^x) \ dx = -\cos (e^x) + C.$

Let $u = e^x$

Compute $du$

$\int e^x\cos (e^x) \ dx = \answer {\sin (e^x)} +C.$

Let $u = \ln (x)$

Compute $du$

$\int \frac {\cos (\ln (x))}{x} \ dx = \answer {\sin (\ln (x))} +C.$

Compute $\int 3x^2\sec (x^3)\tan (x^3) \ dx.$ Let $u = x^3$ . Then we have, $u = x^3$ $du = 3x^2 \ dx$

and the integral can be written as

$\begin{align*} \int 3x^2\sec(x^3)\tan(x^3) \ dx &= \int \sec(x^3)\tan(x^3) \cdot 3x^2 \ dx\\ &= \int \sec(u)\tan(u) \ du. \end{align*}$

The last integral can be computed as $\int \sec (u)\tan (u) \ du = \sec (u) + C$ and by back substituting, we have $\sec (u) + C = \sec (x^3) + C.$ Thus, using u-substitution we can conclude that $\int 3x^2\sec (x^3)\tan (x^3) \ dx = \sec (x^3) + C.$

Let $u = e^x$

Compute $du$

$\int e^x\sec (e^x)\tan (e^x) \ dx = \answer {\sec (e^x)} +C.$

Let $u = x^2$

Compute $du$

$\int 2x\csc (x^2)\cot (x^2) \ dx = \answer {-\csc (x^2)} +C.$

Compute $\int x^3\cos (x^4) \ dx.$ Let $u = x^4$ . Then, $u = x^4$ $du = 4x^3 \ dx$

and the integral can be written as

$\begin{align*} \int x^3\cos(x^4) \ dx &= \int \cos(x^4) \cdot x^3\ dx \\ &= \int \cos(x^4)\cdot \tfrac14 \cdot 4x^3\ dx\\ &= \int \tfrac14\cos(u) \ du. \end{align*}$

The last integral can be computed as $\int \tfrac 14 \cos (u) \ du = \tfrac 14 \sin (u) + C$ and by back substituting, we have $\tfrac 14 \sin (u) + C = \tfrac 14 \sin (x^4) + C.$ Thus, using u-substitution we can conclude that $\int x^3\cos (x^4) \ dx = \tfrac 14 \sin (x^4) + C.$

Let $u = x^2$

Compute $du$

$\int x\cos (x^2) \ dx = \answer {\sin (x^2)/2} +C.$

Compute $\int xe^{-x^2} \ dx.$ Let $u = -x^2$ . Then we have, $u = -x^2$ $du = -2x \ dx$ and the integral can be written as $\begin{align*} \int xe^{-x^2} \ dx &= \int e^{-x^2} \cdot x\ dx \\ &= \int e^{-x^2}( -\tfrac12)\ (-2x)\ dx \\ &= \int -\tfrac12 e^u \ du. \end{align*}$

The last integral can be computed as $\int -\tfrac 12 e^u \ du = -\tfrac 12 e^u + C$ and by back substituting, we have $-\tfrac 12 e^u + C = -\tfrac 12 e^{-x^2} + C.$ Thus, using u-substitution we can conclude that $\int xe^{-x^2} \ dx = -\tfrac 12 e^{-x^2} + C.$

Let $u = x^3$

Compute $du$

$\int x^2e^{x^3} \ dx = \answer {e^{x^3}/3} +C.$

Let $u = \sqrt x$

Compute $du$

$\int \frac {e^{\sqrt x}}{\sqrt x} \ dx = \answer {2e^{\sqrt x}} +C.$

Compute $\int \sin (3x) \ dx.$ Let $u = 3x$ . Then we have, $u = 3x$ $du = 3 \ dx$ and the integral can be written as $\int \sin (3x) \ dx = \int \sin (3x) \cdot \tfrac 13\cdot 3 \ dx = \int \tfrac 13 \sin (u) \ du.$ The last integral can be computed as $\int \tfrac 13 \sin (u) \ du = -\tfrac 13 \cos (u) + C$ and by back substituting, we have $-\tfrac 13 \cos (u) + C = -\tfrac 13 \cos (3x) + C.$ Thus, using u-substitution we can conclude that $\int \sin (3x) \ dx = -\tfrac 13 \cos (3x) + C.$

Compute $\int e^{\frac {x}{2}} \ dx.$ Let $u = \frac {x}{2}$ . Then we have, $u = \frac {x}{2}$ $du = \tfrac 12 \ dx$ and the integral can be written as $\int e^{\frac {x}{2}} \ dx = \int e^{\frac {x}{2}} \cdot 2\cdot \tfrac {1}{2} \ dx = \int 2e^u \ du.$ The last integral can be computed as $\int 2e^u \ du = 2 e^u + C$ and by back substituting, we have $2e^u + C = 2e^{\frac {x}{2}}+ C.$ Thus, using u-substitution we can conclude that $\int e^{x/2} \ dx = 2e^{x/2} + C.$

Compute $\int (4x+3)^5 \ dx.$ Let $u = 4x+3$ . Then we have, $u = 4x+3$ $du = 4 \ dx$ and the integral can be written as $\int (4x+3)^5 \ dx = \int (4x+3)^5 \cdot \tfrac 14\cdot 4 \ dx = \int \tfrac 14 u^5 \ du.$ The last integral can be computed as $\tfrac 14 \int u^5 \ du = \tfrac 14 \cdot \tfrac {u^6}{6} + C = \tfrac {u^6}{24} + C$ and by back substituting, we have $\tfrac {u^6}{24} + C = \tfrac {(4x+3)^6}{24}+ C.$ Thus, using u-substitution we can conclude that $\int (4x+3)^5 \ dx = \tfrac {1}{24}(4x+3)^6 + C.$

Compute $\int \frac {1}{x\ln ^2(x)} \ dx.$ Let $u = \ln (x)$ . Then we have, $u = \ln (x)$ $du = \tfrac 1x \ dx$ and the integral can be written as $\int \frac {1}{x\ln ^2(x)} \ dx = \int \frac {1}{\ln ^2(x)} \cdot \frac {1}{x}\ dx = \int \frac {1}{u^2} \ du.$ The last integral can be computed as $\int \frac {1}{u^2} \ du = \int u^{-2} \ du = \tfrac {u^{-1}}{-1} + C = -\frac {1}{u} + C$ and by back substituting, we have $-\frac {1}{u} + C = -\frac {1}{\ln (x)} + C.$ Thus, using u-substitution we can conclude that $\int \frac {1}{x\ln ^2(x)} \ dx = -\frac {1}{\ln (x)} + C.$

Let $u = \ln (x)$

Compute $du$

$\int \frac {1}{x\ln ^3(x)} \ dx = \answer {- \ln ^{-2}(x)/2} +C.$

Compute $\int \sin ^4(x)\cos (x) \ dx.$ Let $u = \sin (x)$ . Then we have, $u = \sin (x)$ $du = \cos (x) \ dx$ and the integral can be written as $\int \sin ^4(x)\cos (x) \ dx = \int u^4 \ du.$ The last integral can be computed as $\int u^4 \ du = \tfrac {u^5}{5} + C$ and by back substituting, we have $\tfrac {u^5}{5} + C = \tfrac 15 \sin ^5(x) + C.$ Thus, using u-substitution we can conclude that $\int \sin ^4(x)\cos (x) \ dx = \tfrac 15 \sin ^5(x) + C.$

Let $u = \cos (x)$

Compute $du$

$\int \cos ^3(x)\sin (x) \ dx = \answer {-\cos ^4(x)/4} +C.$

Let $u = \tan (x)$

Compute $du$

$\int \tan ^5(x)\sec ^2(x) \ dx = \answer {\tan ^6(x)/6} +C.$

Compute $\int \tan (x) \ dx.$ First rewrite the integral: $\int \tan (x) \ dx =\int \frac {\sin (x)}{\cos (x)} \ dx.$ Now, let $u = \cos (x)$ ; then $du = -\sin (x) \ dx$ and the integral can be written as $\begin{align*} \int \frac{\sin(x)}{\cos(x)} \ dx &= \int \frac{1}{\cos(x)}\ \sin(x) \ dx \\ &= - \int \frac{1}{\cos(x)}\ \big(-\sin(x)\big) \ dx\\ &=-\int \frac{1}{u} \ du. \end{align*}$

The last integral can be computed as $-\int \frac {1}{u} \ du = -\ln |u| + C$ and by back substituting, we have $-\ln |u| + C = -\ln |\cos (x)| + C = \ln |\sec (x)| +C.$ Thus, using u-substitution we can conclude that $\int \tan (x) \ dx = \ln |\sec (x)| + C.$

Rewrite: $\cot (x) = \frac {\cos (x)}{\sin (x)}$

Let $u = \sin (x)$

Compute $du$

$\int \cot (x) \ dx = \answer {\ln |\sin (x)|} +C.$

Compute $\int \sec (x) \ dx.$ First rewrite the integral: $\int \sec (x) \ dx =\int \sec (x)\frac {\sec (x)+\tan (x)}{\sec (x)+\tan (x)} \ dx.$ Distributing in the numerator, we get $\int \sec (x)\frac {\sec (x)+\tan (x)}{\sec (x)+\tan (x)} \ dx = \int \frac {\sec ^2(x)+\sec (x)\tan (x)}{\sec (x)+\tan (x)} \ dx.$ Now, let $u = \sec (x) + \tan (x)$ , then $du = [\sec (x)\tan (x) + \sec ^2(x)] \ dx$ and the integral can be rewritten as $\int \frac {\sec ^2(x)+\sec (x)\tan (x)}{\sec (x)+\tan (x)} \ dx = \int \frac {1}{u} \ du.$ The last integral can be computed as $\int \frac {1}{u} \ du = \ln |u| + C$ and by back substituting, we have $\ln |u| + C = \ln |\sec (x) + \tan (x)| + C.$ Thus, using u-substitution we can conclude that $\int \sec (x) \ dx = \ln |\sec (x) + \tan (x)| + C.$

Here is a detailed, lecture style video on u-substitution: