# Integration By Substitution

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Integration by Substitution

Integration by substitution is another way to reverse the chain rule. In this one, we replace the integration variable $x$ with a different variable $u=f(x)$. We must also replace $dx$ with $du=f'(x)dx$ and replace the limits of the integral too. The aim is to end up with an integral that is easier to evaluate.

A Level

## 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 $x$.

A Level

## 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}=0.0317\end{aligned}

A Level

## 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}}du\end{aligned}

Putting it in the integral:

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

A Level

## 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}\int\left(\dfrac{\sec(x)}{\tan(x)}\right)^{8}dx&=\int\dfrac{\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}}du\end{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}&\int\left(\dfrac{\sec(x)}{\tan(x)}\right)^{8}dx=\int\dfrac{(u^{2}+1)^{3}}{u^{8}}du\\[1.2em]&=\int\dfrac{u^{6}+3u^{4}+3u^{2}+1}{u^{8}}du\\[1.2em]&=\int \left( u^{-2}+3u^{-4}+3u^{-6}+u^{-8}du\right) \\[1.2em]&=-u^{-1}-\left( 3\times\dfrac{1}{3}u^{-3}\right) -\left( 3\times\dfrac{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)+c\end{aligned}

A Level

## Example Questions

Choose $u=x^{4}$

$\dfrac{du}{dx}=4x^{3}$

$du=4x^{3}dx$

$dx=\dfrac{1}{4x^{3}}du$

Put into integral:

\begin{aligned}\int x^{3}e^{x^{4}}dx&=x^{3}e^{u}\dfrac{1}{4x^{3}}du\\[1.2em]&=\int\dfrac{1}{4}e^{u}du\\[1.2em]&=\dfrac{1}{4}e^{u}+c\\[1.2em]&=\dfrac{1}{4}e^{x^{4}}+c\end{aligned}
$u=\sin(x)$

$\dfrac{du}{dx}=\cos(x)$

$du=\cos(x)dx$

$dx=\dfrac{1}{\cos(x)}du$

Lower limit $x=0$:

$u=\sin(0)$

$u=0$

Upper limit $x=\dfrac{\pi}{2}$

$u=\sin(\dfrac{\pi}{2})$

$u=1$

Put into integral:

\begin{aligned}\int^{\frac{\pi}{2}}_{0}\cos(x)sin^{2}(x)dx&=\int^{1}_{0}\cos(x)u^{2}\dfrac{1}{\cos(x)}du\\[1.2em]&=\int^{1}_{0}u^{2}du\\[1.2em]&=\left[\dfrac{1}{3}u^{3}\right]^{1}_{0}=\dfrac{1}{3}\times1^{3}-\dfrac{1}{3}\times0^{3}\\[1.2em]&=\dfrac{1}{3}\end{aligned}

Choose $u=2x^{5}-16$

$\dfrac{du}{dx}=10x^{4}$

$du=10x^{4}dx$

$dx=\dfrac{1}{10x^{4}}du$

Put it into the integral:

\begin{aligned}\int\dfrac{10x^{4}}{(2x^{5}-16)^{3}}dx&=\int\dfrac{10x^{4}}{u^{3}}\dfrac{1}{10x^{4}}du\\[1.2em]&=\int\dfrac{1}{u^{3}}du\\[1.2em]&=\int u^{-3}du\\[1.2em]&=-\dfrac{1}{2}u^{-2}+c\\[1.2em]&=-\dfrac{1}{2}(2x^{5}-16)^{-2}+c\end{aligned}
$u=\cot(x)$

$\dfrac{du}{dx}=-\cosec^{2}(x)$

$du=-\cosec^{2}(x)dx$

$dx=\dfrac{-1}{\cosec^{2}(x)}du$

Put it into the integral:

\begin{aligned}\int\dfrac{\cosec^{4}(x)}{\cot^{\frac{1}{3}}(x)}dx&=\int\dfrac{\cosec^{4}(x)}{u^{\frac{1}{3}}}\dfrac{(-1)}{\cosec^{2}(x)}du\\[1.2em]&=\int-\dfrac{\cosec^{2}(x)}{u^{\frac{1}{3}}}du\end{aligned}

Recall: $\cosec^{2}(x)=1+\cot^{2}(x)$

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

\begin{aligned}\int\dfrac{\cosec^{4}(x)}{\cot^{\frac{1}{3}}(x)}dx&=\int-\dfrac{1+u^{2}}{u^{\frac{1}{3}}}du\\[1.2em]&=\int-u^{-\frac{1}{3}}-u^{\frac{5}{3}}du\\[1.2em]&=-\dfrac{3}{2}u^{\frac{2}{3}}-\dfrac{3}{8}u^{\frac{8}{3}}+c\\[1.2em]&=-\dfrac{3}{2}\cot^{\frac{2}{3}}(x)-\dfrac{3}{8}\cot^{\frac{8}{3}}(x)+c\end{aligned}

A Level

A Level

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