Integration By Substitution
Integration By Substitution Revision
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.
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}=0.0317\end{aligned}
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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}
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}
Integration By Substitution Example Questions
Question 1: Use a suitable substitution to evaluate \int x^{3}e^{x^{4}}dx.
[4 marks]
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}
Question 2: Evaluate \int^{\frac{\pi}{2}}_{0}\cos(x)\sin^{2}(x)dx by using the substitution u=\sin(x).
[4 marks]
\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}
Question 3: Find {\LARGE \int}\dfrac{10x^{4}}{(2x^{5}-16)^{3}}dx
[4 marks]
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}
Question 4: Using the substitution u=\cot(x), find {\LARGE \int}\dfrac{\cosec^{4}(x)}{\cot^{\frac{1}{3}}(x)}dx
[6 marks]
\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}
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