How to Do Integration by Substitution

Substitution in integration is the chain rule reversed. The formula is as follows:

Formula

Substitution

$\begin{array}{llll}\hfill \int f\left(g\left(x\right)\right)\cdot {g}^{\prime }\left(x\right)\phantom{\rule{0.17em}{0ex}}dx& =\int f\left(u\right)\phantom{\rule{0.17em}{0ex}}du\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =F\left(u\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =F\left(g\left(x\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You use substitution when the expression contains two functions where one of the functions is the derivative of the other. Choose $u$ as the function that is not differentiated! Next, you differentiate both sides with respect to $x$ so that you have $\frac{\phantom{\rule{0.17em}{0ex}}du}{\phantom{\rule{0.17em}{0ex}}dx}$. Then, you multiply by $\phantom{\rule{0.17em}{0ex}}dx$ on both sides of the equation to cancel the $\phantom{\rule{0.17em}{0ex}}dx$ in the denominator, such that you have solved for $\phantom{\rule{0.17em}{0ex}}du$.

Example 1

$\begin{array}{llll}\hfill \int 2x{e}^{{x}^{2}}\phantom{\rule{0.17em}{0ex}}dx& \stackrel{\ast }{=}\int {e}^{u}\phantom{\rule{0.17em}{0ex}}du\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{u}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{{x}^{2}}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\int 2x{e}^{{x}^{2}}\phantom{\rule{0.17em}{0ex}}dx\stackrel{\ast }{=}\int {e}^{u}\phantom{\rule{0.17em}{0ex}}du={e}^{u}+C={e}^{{x}^{2}}+C$

*

$\begin{array}{ccc}\hfill u& ={x}^{2}\hfill & \hfill \\ \hfill \frac{\phantom{\rule{0.17em}{0ex}}du}{\phantom{\rule{0.17em}{0ex}}dx}& =2x\hfill \\ \hfill \phantom{\rule{0.17em}{0ex}}du& =2x\phantom{\rule{0.17em}{0ex}}dx\hfill \end{array}$

Example 2

Compute $\int \frac{2x+1}{{x}^{2}+x}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill \int \frac{2x+1}{{x}^{2}+x}\phantom{\rule{0.17em}{0ex}}dx& \stackrel{\ast }{=}\int \frac{1}{u}\phantom{\rule{0.17em}{0ex}}du\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|u|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|{x}^{2}+x|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

*

$\begin{array}{ccc}\hfill u& ={x}^{2}+x\hfill & \hfill \\ \hfill \frac{\phantom{\rule{0.17em}{0ex}}du}{\phantom{\rule{0.17em}{0ex}}dx}& =2x+1\hfill \\ \hfill \phantom{\rule{0.17em}{0ex}}du& =2x+1\phantom{\rule{0.17em}{0ex}}dx\hfill \end{array}$

Example 3

Compute $\int \frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill \int \frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}\phantom{\rule{0.17em}{0ex}}dx& \stackrel{\ast }{=}\int -\frac{1}{{u}^{2}}\phantom{\rule{0.17em}{0ex}}du\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\int {u}^{-2}\phantom{\rule{0.17em}{0ex}}du\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={u}^{-1}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{u}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{\mathrm{cos}x}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

*

$\begin{array}{ccc}\hfill u& =\mathrm{cos}x\hfill & \hfill \\ \hfill \frac{\phantom{\rule{0.17em}{0ex}}du}{\phantom{\rule{0.17em}{0ex}}dx}& =-\mathrm{sin}x\hfill \\ \hfill \phantom{\rule{0.17em}{0ex}}du& =-\mathrm{sin}x\phantom{\rule{0.17em}{0ex}}dx\hfill \end{array}$

Example 4

Compute $\int \mathrm{cos}x{e}^{\mathrm{sin}x}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill \int \mathrm{cos}x{e}^{\mathrm{sin}x}\phantom{\rule{0.17em}{0ex}}dx& \stackrel{\ast }{=}\int {e}^{u}\phantom{\rule{0.17em}{0ex}}du\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{u}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{\mathrm{sin}x}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

*

$\begin{array}{ccc}\hfill u& =\mathrm{sin}x\hfill & \hfill \\ \hfill \frac{\phantom{\rule{0.17em}{0ex}}du}{\phantom{\rule{0.17em}{0ex}}dx}& =\mathrm{cos}x\hfill \\ \hfill \phantom{\rule{0.17em}{0ex}}du& =\mathrm{cos}x\phantom{\rule{0.17em}{0ex}}dx\hfill \end{array}$