7.4 Integration by Substitution

Let $u=x^2+3$, thus $du=2xdx,$ or $\frac{1}{2}du=xdx$. Now we can rewrite the integral as $$\begin{array}{rl}{\displaystyle \int \underset{u^{17}}{\underbrace{(x^2+3{)}^{17}}}\underset{\frac{1}{2}du}{\underbrace{xdx}}}& =\int \frac{1}{2}{u}^{17}du\\ & =\frac{1}{2}\frac{u^{18}}{18}+C\\ & =\frac{1}{36}(x^2+3{)}^{18}+C.\end{array}$$

Let $u=x^3+1$. Then $du=3x^{2}dx$ or $x^{2}dx=\frac{1}{3}du$, and $$\begin{array}{rl}\int x^2\sqrt{x^3+1}dx& =\int \sqrt{\underset{=u}{\underbrace{x^3+1}}}\stackrel{\frac{1}{3}du}{\overbrace{x^{2}dx}}\\ & =\int \frac{1}{3}\sqrt{u}du\\ \text{(}{u}^{1/2}=\sqrt{u}\text{)}& & =\frac{1}{3}\int u^{1/2}du\text{(}\int u^{\alpha }du=\frac{1}{\alpha +1}{u}^{\alpha +1}+C\text{)}& & =\frac{1}{3}\left(\frac{2}{3}{u}^{3/2}\right)+C\text{(}u=x^3+1\text{)}& & =\frac{2}{9}(x^3+1{)}^{3/2}+C.\end{array}$$

Let $u=ax$. Then $du=adx$ and $$\begin{array}{rl}\int \sin axdx& =\frac{1}{a}\int \sin udu\\ & =-\frac{1}{a}\cos u+C\\ & =-\frac{1}{a}\cos ax+C.\end{array}$$ Similarly $$\int \cos axdx=\frac{1}{a}\sin ax+C.$$

A simple substitution is so useful that is worth noting explicitly.

(a) Because $\int {\sec }^{2}xdx=\tan x+C$, it follows from the above theorem that $$\int {\sec }^2(x-1)dx=\tan (x-1)+C.$$ (b) Since $\int \sin xdx=-\cos x+C$, the above theorem gives $$\int \sin (2x-\pi /4)dx=-\frac{1}{2}\cos (2x-\pi /4)+C$$

Let $u=\cos x$. Then $$du=-\sin xdx$$ and $$\begin{array}{rl}\int {\cos }^{2}x\sin xdx& =-\int u^{2}du\\ & =-\frac{u^3}{3}+C\\ & =-\frac{1}{3}{\cos }^{3}x+C.\end{array}$$

Recall that $\tan x=\sin x/\cos x$. Let $u=\cos x$. Then $$du=-\sin xdx$$ and $$\begin{array}{rl}\int \tan xdx& =\int \frac{\sin x}{\cos x}dx\\ & =\int \frac{-du}{u}\\ & =-\ln |u|+C\\ & =-\ln |\cos x|+C.\end{array}$$ Recall that $\ln x^{\alpha }=\alpha \ln x$. So $$\begin{array}{rl}-\ln |\cos x|+C& =\ln (|\cos x{|}^{-1})+C\\ & =\ln |\sec x|+C.\end{array}$$ Therefore, the result can also be written as $$\int \tan xdx=\ln |\sec x|+C.$$

$$\int \frac{1}{a^2+x^2}dx=\frac{1}{a^2}\int \frac{1}{1+{\left(\frac{x}{a}\right)}^2}dx.$$ Let $u=x/a$. Then $$dx=adu$$ and $$\begin{array}{rl}\frac{1}{a^2}\int \frac{1}{1+(x/a{)}^2}dx& =\frac{1}{a^2}\int \frac{1}{1+u^2}adu\\ & =\frac{1}{a}\arctan u+C\\ & =\frac{1}{a}\arctan \frac{x}{a}+C.\end{array}$$ To evaluate $\int \frac{1}{1+(x/a{)}^2}dx$, we could simply use Theorem above.

By completing the square in the denominator, we have $$\begin{array}{rl}\int \frac{1}{x^2+2x+5}dx& =\int \frac{dx}{(x^2+2x+1)+4}\\ & =\int \frac{dx}{(x+2{)}^2+2^2}.\end{array}$$ Now let $u=x+2$. Then $du=dx$ and $$\begin{array}{rl}\int \frac{dx}{(x+2{)}^2+1}& =\int \frac{du}{u^2+4}\\ \text{(from previous example)}& & =\frac{1}{2}\arctan \frac{u}{2}+C& =\frac{1}{2}\arctan \frac{x+2}{2}+C.\end{array}$$

Let $a^x=e^u$. Taking the natural logarithm of each side, we get $$\begin{array}{rl}\ln (a^x)& =\ln e^u\\ x\ln a& =u.\end{array}$$ Thus $$x\ln a=u\Rightarrow \ln adx=du.$$ Now we can write $$\begin{array}{rl}\int a^{x}dx& =\int e^u\frac{du}{\ln a}\\ & =\frac{1}{\ln a}{e}^u+C\\ & =\frac{1}{\ln a}{a}^x+C.\end{array}$$

We can directly use Theorem above or use the substitution $u=ax$. The substitution leads to $$du=adx$$ and $$\begin{array}{rl}\int e^{ax}dx& =\int e^u\frac{du}{a}\\ & =\frac{1}{a}\int e^{u}du\\ & =\frac{1}{a}{e}^u+C\\ & =\frac{1}{a}{e}^{ax}+C.\end{array}$$

Note that the substitution $u=1+e^{2x}$ does not work because $du=2e^{2x}dx(\ne e^{x}dx)$ does not appear in the numerator. But if we notice that $1+e^{2x}=1+(e^x{)}^2$, we can make the subtitution $u=e^x$ to make the denominator of the form $1+u^2$. Then $$u=e^x\Rightarrow du=e^{x}dx$$ and $$\begin{array}{rl}\int \frac{e^x}{1+e^{2x}}dx& =\int \frac{du}{1+u^2}\\ & =\arctan u+C\\ \text{(}u=e^x\text{)}& & =\arctan e^x+C.\end{array}$$

Because $(\cosh x{)}^{\prime }=\sinh x$, we know that $$\int \sinh xdx=\cosh x+C.$$ Also we can use the definition of $\sinh x$ for evaluation of the integral: $$\sinh x=\frac{e^x-e^{-x}}{2}$$ [Recall that $\sinh x$ (similar to $\sin x$) is an odd function, so it is half of $e^x-e^{-x}$ and $\cosh x$ (similar to $\cos x$) is an even function, so it is half of $e^x+e^{-x}$.] $$\begin{array}{rl}\int \sinh xdx& =\frac{1}{2}\int (e^x-e^{-x})dx\\ & =\frac{1}{2}[e^x-(-e^{-x})]+C\\ & =\cosh x+C\end{array}$$ Similarly we can show $$\int \cosh xdx=\sinh x+C.$$

$$\int \tanh xdx=\int \frac{\sinh x}{\cosh x}dx.$$ Let $u=\cosh x$. Then $du=\sinh xdx$ and $$\begin{array}{rl}\int \frac{\sinh x}{\cosh x}dx& =\int \frac{du}{u}\\ & =\ln |u|+C\\ & =\ln |\cosh x|+C.\end{array}$$ Because $\cosh x=(e^x+e^{-x})/2>0$, so $|\cosh x|=\cosh x$ and $$\int \tanh xdx=\ln (\cosh x)+C.$$

Note that $\sqrt[x]{7}=7^{1/x}.$ If we let $\frac{1}{x}=u$, because $$\frac{1}{x}=u\Rightarrow -\frac{1}{x^2}dx=du,$$ we have $$\int \frac{\sqrt[x]{7}}{x^2}dx=-\int 7^{u}du=-\frac{1}{\ln 7}{7}^u+C=-\frac{1}{\ln 7}\sqrt[x]{7}+C.$$

Let $u=\ln x$, then $$du=\frac{1}{x}dx$$ and $$\begin{array}{rl}\int \underset{u}{\underbrace{\frac{1}{\ln x}}}\underset{du}{\underbrace{\frac{1}{x}dx}}& =\int \frac{du}{u}\\ & =\ln |u|+C\\ \text{(}u=\ln x\text{)}& & =\ln |\ln x|+C\end{array}$$

Let $u=\ln x$, then $$du=\frac{1}{x}dx.$$ So $$\begin{array}{rl}\int \frac{\sin (\ln x)}{x}dx& =\int \sin udu\\ & =-\cos u+C\\ & =-\cos (\ln x)+C.\end{array}$$

Let $u=\arctan x^2$, then $$\begin{array}{rl}du& =2x\frac{1}{1+(x^2{)}^2}dx\\ & =\frac{2x}{1+x^4}dx.\end{array}$$ Therefore $$\begin{array}{rl}\int \frac{x}{1+x^4}\arctan x^{2}dx& =\int \frac{1}{2}\underset{u}{\underbrace{\arctan x^2}}\underset{du}{\underbrace{\frac{2x}{1+x^4}dx}}\\ & =\frac{1}{4}{u}^2+C\\ & =\frac{1}{4}{\arctan }^2{x}^2+C.\end{array}$$