Integration By Substitution

How to Integrate by Substitution

Step 1: You will be presented with an integrand that is made up of two functions of $x$.

Step 2: Substitute $u=f(x)$ where $f(x)$ is one of the functions of $x$.

Step 3: Find $dfrac{du}{dx}$ then rearrange to get $dx$ in terms of $du$.

Step 4: Rewrite the original integral in terms of $u$ and $du$ and simplify it.

Step 5: If you chose your substitution well, you will now be left with something much easier to integrate.

Step 6: Integrate it.

Step 7: Substitute $u$ for $f(x)$ in the answer to get the final answer in terms of $x$.

Changing the Limits

For a definite integral of the form $int^{b}_{a}$, if our substitution is $u=f(x)$, then rather than substitute $x$ back in at the end, we can change the limits to $int^{f(b)}_{f(a)}$ and put those limits into our expression for $u$ to evaluate the integral.

Example: Find $int^{0.5}_{0}5x^{4}e^{x^{5}}dx$, using the substitution $u=x^{5}$.

$u=f(x)=x^{5}$

So limits become:

$0.5^{5}=0.03125$ and $0^{5}=0$

$begin{aligned}dfrac{du}{dx}=5x^{4}[1.2em]du=5x^{4}dx[1.2em]dx=dfrac{du}{5x^{4}}end{aligned}$

Integral becomes:

$begin{aligned}int^{0.03125}_{0}5x^{4}e^{u}dfrac{du}{5x^{4}}&=int^{0.03125}_{0}e^{u}du[1.2em]&=[e^{u}]^{0.03125}_{0}[1.2em]&=e^{0.03125}-e^{0}=0.0317end{aligned}$

Integration by Substitution on Fractions

When choosing a substitution for a fraction, the best thing to choose is almost always the denominator or part of the denominator.

Example: Integrate ${LARGE int}dfrac{4x^{3}}{(x^{4}-1)^{frac{1}{6}}}dx$ with a suitable substitution.

Choose $u=x^{4}-1$.

$begin{aligned}dfrac{du}{dx}=4x^{3}[1.2em]du=4x^{3}dx[1.2em]dx=dfrac{1}{4x^{3}}duend{aligned}$

Putting it in the integral:

$begin{aligned}intdfrac{4x^{3}}{(x^{4}-1)^{frac{1}{6}}}dx&=intdfrac{4x^{3}}{u^{frac{1}{6}}}dfrac{1}{4x^{3}}du[1.2em]&=intdfrac{1}{u^{frac{1}{6}}}du[1.2em]&=int u^{-frac{1}{6}}du[1.2em]&=dfrac{6}{5}u^{frac{5}{6}}+c[1.2em]&=dfrac{6}{5}(x^{4}-1)^{frac{5}{6}}+cend{aligned}$

Trigonometric Integration by Substitution

Integration by substitution questions involving trigonometry can be very difficult. They involve not only the skills on this page, but also a good knowledge of trigonometric integration and trigonometric identities is a must.

Example: Integrate $left(dfrac{sec(x)}{tan(x)}right)^{8}$ using the substitution $u=tan(x)$.

$u=tan(x)$

$dfrac{du}{dx}=sec^{2}(x)$

$du=sec^{2}xdx$

$dx=dfrac{1}{sec^{2}(x)}du$

Put into integral:

$begin{aligned}intleft(dfrac{sec(x)}{tan(x)}right)^{8}dx&=intdfrac{sec^{8}(x)}{tan^{8}(x)}dx[1.2em]&=dfrac{sec^{8}(x)}{u^{8}}dfrac{1}{sec^{2}(x)}du[1.2em]&=dfrac{sec^{6}(x)}{u^{8}}duend{aligned}$

How do we deal with the $sec^{6}$ term?

Recall: $sec^{2}(x)=tan^{2}(x)+1$

$sec^{2}(x)=u^{2}+1$

$sec^{6}(x)=(u^{2}+1)^{3}$

$begin{aligned}&intleft(dfrac{sec(x)}{tan(x)}right)^{8}dx=intdfrac{(u^{2}+1)^{3}}{u^{8}}du[1.2em]&=intdfrac{u^{6}+3u^{4}+3u^{2}+1}{u^{8}}du[1.2em]&=int left( u^{-2}+3u^{-4}+3u^{-6}+u^{-8}duright) [1.2em]&=-u^{-1}-left( 3timesdfrac{1}{3}u^{-3}right) -left( 3timesdfrac{1}{5}u^{-5}right) -dfrac{1}{7}u^{-7}+c[1.2em]&=-u^{-1}-u^{-3}-dfrac{3}{5}u^{-5}-dfrac{1}{7}u^{-7}+c[1.2em]&=-cot(x)-cot^{3}(x)-dfrac{3}{5}cot^{5}(x)-dfrac{1}{7}cot^{7}(x)+cend{aligned}$