# Examples On Integration By Substitution Set-1

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A lot many times, we will encounter functions whose integrals cannot be obtained from their original expressions; however, an appropriate substitution might reduce the given function to another function whose integral is obtainable.

This method of integration by substitution is used extensively to evaluate integrals. As we progress along this section we will develop certain rules of thumb that will tell us what substitutions to use where. Also, multiple substitutions might be possible for the same function. Therefore, integration by substitution is more of an art and you can develop the knack of it only by extensive practice (and of course, some thinking !)

Example - 11

Evaluate \begin{align}\int {\frac{{{{\cos }^3}x}}{{{{\sin }^2}x + \sin x}}} \,dx\end{align}

Solution: The general approach while substitution is as follows:

Suppose we have to obtain $$\int {f\left( x \right)dx}$$ . We find some function $$\phi \left( x \right)$$ (and put it equal to a variable y) such that $$\phi '\left( x \right)dx = dy$$ is some part of $$f(x)dx$$. This will let us express $$\int {f\left( x \right)dx}$$ in terms of another integral which contains only y.

This approach will become quite clear when we apply it on the given example:

$I = \int {\frac{{{{\cos }^3}x}}{{{{\sin }^2}x + \sin x}}} \,dx = \int {\frac{{\left( {1 - {{\sin }^2}x} \right)\cos x}}{{{{\sin }^2}x + \sin x}}} \,dx$

Observe carefully why we wrote $${\cos ^2}x$$ in the numerator of $$I$$ as $$\left( {1 - {{\sin }^2}x} \right);$$ if we now substitute $$\sin x = y,$$ we’ll get $$\cos x\,dx = dy,$$ so that the entire expression of $$I$$ can be reduced to another integral which contains only y:

\begin{align}&\sin x = y\\& \cos x\,dx = dy\\ \end{align}

\begin{align}& \,I=\int{\frac{\left( 1-{{y}^{2}} \right)}{{{y}^{2}}+y}}dy \\ & \,\,\,\text{ }\!\!~\!\!\text{ }=\int{\frac{\left( 1-y \right)}{y}}dy \\ & \,\,\,=\int{\left( \frac{1}{y}-1 \right)}dy \\ & \,\,\,=\ln \left| y \right|-y+C \\ \end{align}

We see that the modified integral (the integral in terms of y) was easily integrable; to obtain the integral in terms of x, we now simply substitution $$y=\sin \,x$$:

$I = \ln \left| {\sin x} \right| - \sin x + C$

Example - 12

Evaluate \begin{align}\int {\frac{{{x^7}}}{{{{\left( {1 - {x^2}} \right)}^5}}}} \,\,dx\end{align}.

Solution: This example will serve to show that multiple substitutions are possible for the same function.

(a) Notice that the numerator , $${x^7}dx,$$ can be written as $${x^6} \cdot xdx$$ . If we substitute $${x^2} = y,$$ we’ll obtain $$xdx = \frac{{dy}}{2}$$ so that the entire integral can be expressed in terms of y. However, the integral will become

$z\frac{1}{2}\int {\frac{{{y^3}}}{{{{\left( {1 - y} \right)}^5}}}\,\,dy}$

which still cannot be integrated directly because of the denominator $${\left( {1 - y} \right)^5}$$ . What we therefore do is substitute $$\left( 1-{{x}^{2}} \right)=y\ \text{instead}\ \text{of}\ {{x}^{\text{2}}}\text{=y,}$$ because then the denominator will be reduced further directly:

\begin{align}&\quad\;\;1 - {x^2} = y\\& \Rightarrow \quad xdx = \frac{{ - dy}}{2} \end{align}

\begin{align}& I = \int {\frac{{{x^7}}}{{{{\left( {1 - {x^2}} \right)}^5}}}dx = \int {\frac{{{x^6} \cdot x}}{{{{\left( {1 - {x^2}} \right)}^5}}}\,\,dx} }\\&\qquad\qquad\qquad\qquad= - \frac{1}{2}\int {\frac{{{{\left( {1 - y} \right)}^3}}}{{{y^5}}}} \,\,dy\\&\qquad\qquad\qquad\qquad = \frac{1}{2}\int {\left\{ {\frac{{{y^3} - 1 - 3{y^2} + 3y}}{{{y^5}}}} \right\}} \,\,dy\\&\qquad\qquad\qquad\qquad= \frac{1}{2}\int {\left\{ {{y^{ - 2}} - 3{y^{ - 3}} + 3{y^{ - 4}} - {y^{ - 5}}} \right\}} \,\,dy\\&\qquad\qquad\qquad\qquad = \frac{1}{2}\left\{ { - \frac{1}{y} + \frac{3}{{2{y^2}}} - \frac{1}{{{y^3}}} + \frac{1}{{4{y^4}}}} \right\}\, + C\\ &\qquad\qquad\qquad\qquad= \frac{{1 - 4y + 6{y^2} - 4{y^3}}}{{8{y^4}}}\, + C\\& \qquad\qquad\quad\Rightarrow \;\;\; 1 = \frac{{1 - 4\left( {1 - {x^2}} \right) + 6{{\left( {1 - {x^2}} \right)}^2} - 4{{\left( {1 - {x^2}} \right)}^3}}}{{8{{\left( {1 - {x^2}} \right)}^2}}}\, + C\end{align}

(b) The denominator contains the term $$\left( {1{\rm{ }}-{x^2}} \right)$$ . Think of a substitution that could cause the denominator to reduce to a single term: this substitution should be trignometric:

\begin{align}&\;x = \sin \theta \\& dx = \cos \,\theta \,d\,\theta \end{align}

\begin{align}&I = \int {\frac{{{x^7}}}{{{{\left( {1 - {x^2}} \right)}^5}}}dx = \int {\frac{{{{\sin }^7}\theta \cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^5}}}\,\,d\theta } } \\&\qquad\qquad\qquad\qquad= \int {\frac{{{{\sin }^7}\theta \cos \theta }}{{{{\cos }^{10}}\theta }}\,\,d\theta } \\&\qquad\qquad\qquad\qquad= \int {{{\tan }^7}\theta {{\sec }^2}\theta \,d\theta } \end{align}

Notice now that the simple substitution $$\tan \theta = y$$  will reduce  $$I$$  to a simple integrable form:

\begin{align}&\qquad \quad\;\;\tan \theta = y\\ &\Rightarrow \quad {\sec ^2}\theta d\theta = dy\end{align}

\begin{align} &I = \int {{{\tan }^7}\theta } {\sec ^2}\theta d\theta = \int {{y^7}dy} \\&\qquad\qquad\qquad\qquad\;= \frac{{{y^8}}}{8} + C \\ &\qquad\qquad\qquad\qquad\;= \frac{{{{\tan }^8}\theta }}{8} + C \\ &\qquad\qquad\qquad\qquad\;= \frac{{{x^8}}}{{8{{\left( {1 - {x^2}} \right)}^4}}} + C\;\;\;\;\;\;\;\;\left( {∵ x = \sin \theta } \right) \end{align}

There is one last point to be observed. The answers obtained by methods (a) and (b) might seem to be different from each other. However, verify that they are not! The answer obtained by (a) can be converted into the answer obtained by (b) by just a simple splitting of the constant of integration.

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