u-Substitution

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Calculus

Cegep/2

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719 words

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5 minutes

Method of reversing an application of chain rule to find the antiderivative

To find the antiderivative of $f(x)=\int g(A)u^{\prime }dx$,

1. Let $u=A$.
2. Find $\frac{\mathrm{d}u}{\mathrm{d}x}$.
3. Substitute $u$ into $A$ and $du$ into $u^{\prime }dx\to f(x)=\int g(u)du$.
4. Integrate $\to G(u)+C$.
5. Back-substitute $A$ into $u\to G(A)+C$.

The limits of integration must be adjusted when using substitution on a definite integral.

Examples

Evaluate the following integrals.

$\int 2x(x^2-1{)}^{4}dx$

Let $u=x^2-1$, then $du=2xdx$.
Using substitution, we get $\int u^{4}du$.
Finding the antiderivative gives $\frac{u^5}{5}+C$.
Using back-substitution, we get $\frac{(x^2-1{)}^5}{5}+C$.

$\int e^{-x^4}(-4x^3)dx$

Let $u=-x^4$, then $du=-4x^{3}dx$.

$$\begin{array}{rl}\int e^{-x^4}(-4x^3)dx& =\int e^{u}du\\ ⟹& =e^u\\ & =e^{-x^4}+C\end{array}$$

$\int \sqrt{\cot x}{\csc }^{2}xdx$

Let $u=\cot x$, then $du=-{\csc }^{2}x\cdot dx$.

$$\begin{array}{rl}\int \sqrt{\cot x}{\csc }^{2}xdx& =-\int \sqrt{u}du\\ ⟹& =-\frac{2u^{\frac{3}{2}}}{3}+C\\ & =-\frac{2(\cot x{)}^{\frac{3}{2}}}{3}+C\end{array}$$

$\int e^{\tan 2x}{\sec }^{2}2xdx$

Let $u=\tan 2x$, then $du={\sec }^{2}2x\cdot 2dx$.

$$\begin{array}{rl}\int e^{\tan 2x}{\sec }^{2}2xdx& =\frac{1}{2}\int e^{u}du\\ ⟹& =\frac{1}{2}{e}^u\\ & =\frac{1}{2}{e}^{\tan 2x}+C\end{array}$$

$\int \frac{2x+3}{(2x^2+6x+5{)}^{\frac{3}{2}}}dx$

Let $u=2x^2+6x+5$, then

$$\begin{array}{rl}du& =(4x+6)dx\\ \frac{du}{2}& =(2x+3)dx\end{array}$$

Using substitution,

$$\begin{array}{rl}\int \frac{2x+3}{(2x^2+6x+5{)}^{\frac{3}{2}}}dx& =\frac{1}{2}\int \frac{du}{u^{\frac{3}{2}}}\\ ⟹& =\frac{1}{5}{u}^{\frac{5}{2}}\\ & =\frac{1}{5}(2x^2+6x+5{)}^{\frac{5}{2}}+C\end{array}$$

$\int {\sec }^{3}x\tan xdx$

Let $u=\sec x$, then $du=\sec x\cos xdx$.

$$\begin{array}{rl}\int {\sec }^{3}x\tan xdx& =\int u^{2}du\\ & =\frac{u^3}{3}+C\\ & =\frac{{\sec }^{3}x}{3}+C\end{array}$$

$\int {\sin }^{2}x{\cos }^{2}xdx$

$$\begin{array}{rl}\int {\sin }^{2}x{\cos }^{2}xdx& =\frac{1}{4}\int (1-\cos (2x))(1+\cos (2x))dx\\ & =\frac{1}{4}\int (1-{\cos }^2(2x))dx\\ & =\frac{1}{4}x-\frac{1}{4}\int {\cos }^2(2x)dx\\ & =\frac{1}{4}x-\frac{1}{4}\left(\frac{1}{2}\right)\int (1+\cos (4x))dx\\ & =\frac{1}{4}x-\frac{1}{8}x-\frac{1}{8}\int \cos (4x)dx\\ & =\frac{1}{8}x-\frac{1}{32}\sin (4x)+C\end{array}$$

$\int {\tan }^{2}xd{\sec }^{2}x$

$$\begin{array}{rl}\int {\tan }^{5}x{\sec }^{7}x& =dx\int {\tan }^{4}x{\sec }^6\sec x\tan xdx\\ & =\int ({\sec }^{2}x-1{)}^2{\sec }^{6}x\sec x\tan xdx\end{array}$$

Let $u=\sec x$, then $du=\sec x\tan xdx$.
By substitution,

$$\begin{array}{rl}& =\int (u^2-1{)}^{2}du\\ & =\int (u^4-2u^2+1)du\\ & =\int (u^{10}-2u^8+u^6)du\\ & =\frac{1}{11}{\sec }^{11}x-\frac{2}{9}{\sec }^{9}x+\frac{1}{7}{\sec }^{7}x+C\end{array}$$

$\int {\tan }^{3}xdx$

$$\begin{array}{rl}\int {\tan }^{3}xdx& =\int ({\sec }^{2}x-1)\tan xdx\\ & =\int {\sec }^{2}x\tan xdx-\int \tan xdx\end{array}$$

Let $u=\tan x$, then $du={\sec }^{2}xdx$.
By substitution,

$$\begin{array}{rl}& =\int udu-\ln |\sec x|+C\\ & =\frac{1}{2}{\tan }^{2}x-\ln |\sec x|+C\end{array}$$

$\int \frac{{\tan }^5\theta }{{\cos }^9\theta }\mathrm{d}\theta$

$$\int \frac{{\tan }^5\theta }{{\cos }^9\theta }\mathrm{d}\theta =\int \frac{{\sin }^5\theta }{{\cos }^{14}\theta }\mathrm{d}\theta$$

Let $u=\cos \theta ⟹du=-\sin \theta d\theta$.

$$\begin{array}{rl}& =-\int \frac{(1-u^2{)}^2}{u^{14}}\mathrm{d}u\\ & =-\int u^{-14}-2u^{-12}+u^{-10}\mathrm{d}u\\ & =\frac{u^{-13}}{13}-\frac{2u^{-11}}{11}+\frac{u^{-9}}{9}+C\\ & =\frac{1}{13{\cos }^{13}\theta }-\frac{2}{11{\cos }^{11}\theta }+\frac{1}{9{\cos }^9\theta }+C\end{array}$$

${\int }_2^8{x}^{-1}({\log }_{8}x{)}^2\mathrm{d}x$

Let $u={\log }_{8}x⟹du=\frac{1}{\ln 8x}dx$.

$$\begin{array}{rl}{\int }_2^8{x}^{-1}({\log }_{8}x{)}^2\mathrm{d}x& ={\int }_{{\log }_{8}2}^{{\log }_{8}8}{u}^2\ln 8\mathrm{d}u\\ & =\ln 8[2u{]}_{{\log }_{8}2}^{{\log }_{8}8}\\ & =\ln 8(2{\log }_{8}8-2{\log }_{8}2)\\ & =\frac{4\ln 8}{3}\end{array}$$

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Ajitani Hifumi

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