# Examples On Integration By Substitution Set-6

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## Examples On Integration By Substitution Set 6

Example-20

Evaluate the following integrals:

(a) \begin{align}\int {\sqrt {\frac{{a - x}}{{a + x}}} dx} \end{align}

(b) \begin{align} \int {\sqrt {\frac{{x - a}}{{b - x}}} dx} \end{align}

Solution: We saw in example-14, a particular example of the general expression in part -(a) above.

As in example-18, we will mention here the appropriate substitutions to be used whenever expressions of the forms above are encountered.

 ${{\rm{Expression}}}$ ${{\rm{Can \;be\; reduced \;by \;the \;substitution}}}$ ${(1)}$ ${(2)}$ ${\sqrt {\frac{{a - x}}{{a + x}}} \,\,{\rm{or}}\,\,\sqrt {\frac{{a + x}}{{a - x}}} }$ ${\sqrt {\frac{{x - a}}{{b - x}}} \,\,{\rm{or}}\,\,\sqrt {(x - a)(b - x)} }$ ${x = a\cos 2\theta }$ ${x = a{{\cos }^2}\theta + b{{\sin }^2}\theta }$

It is straightforward to see why these substitutions are appropriate

(a) \begin{align} & Substitute\qquad\qquad\;\ x=a\cos 2\theta \\ & \qquad\qquad\qquad\Rightarrow\quad dx=-2a\sin 2\theta d\theta \\ & \qquad\quad \;\;and\quad\ (a-x)=a(1-\cos 2\theta ) \\ &\qquad\qquad\qquad\qquad\quad\;\; =2a{{\sin }^{2}}\theta \\ & \qquad\quad\; and\quad\ (a+x)=a(1+\cos 2\theta ) \\ & \qquad\qquad\qquad\qquad\quad\;=2a{{\cos }^{2}}\theta \\ &\qquad\qquad\qquad \Rightarrow\quad I=-2a\int{\sqrt{\frac{2a{{\sin }^{2}}\theta }{2a{{\cos }^{2}}\theta }}\sin 2\theta d\theta } \\ &\qquad\qquad\qquad\qquad\quad =-2a\int{\tan \theta \cdot \sin 2\theta d\theta }\quad.......\left( 1 \right) \\ &\qquad\qquad\qquad\qquad\quad =-4a\int{{{\sin }^{2}}\theta d\theta } \\ & \qquad\qquad\qquad\qquad\quad=4a\left\{ {{\cos }^{2}}\theta -1 \right\}d\theta \\ & \qquad\qquad\qquad\qquad\quad=4a\int{\left\{ \frac{1+\cos 2\theta }{2}-1 \right\}d\theta } \\ & \qquad\qquad\qquad\qquad\quad=2a\int{(\cos 2\theta -1)d\theta } \\ & \qquad\qquad\qquad\qquad\quad=a\sin 2\theta -2a\theta +C \\&\qquad\qquad\qquad\qquad\quad=a\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}}-a\;{{\cos }^{-1}}\frac{x}{a}+C \\ & \qquad\qquad\qquad\qquad\quad=\sqrt{{{a}^{2}}-{{x}^{2}}}-a\;{{\cos }^{-1}}\frac{x}{a}+C \\ \end{align}

(b) \begin{align}& Substitute\qquad\qquad\ x=a\;{{\cos }^{2}}\theta +b\;{{\sin }^{2}}\theta \\ & \qquad\qquad\quad\;\;\;\Rightarrow \quad dx=(b-a)\sin 2\theta d\theta \\ &\qquad\quad\;\;\ and\quad(x-a)=(b-a){{\sin }^{2}}\theta \\ &\qquad\quad\;\; \ and\quad(b-x)=(b-a){{\cos }^{2}}\theta \\ & \qquad\qquad\qquad\Rightarrow\quad I=(b-a)\int{\sqrt{\frac{(b-a){{\sin }^{2}}\theta }{(b-a){{\cos }^{2}}\theta }}\sin 2\theta d\theta } \\ &\qquad\qquad\qquad\qquad\quad =(b-a)\int{\tan \theta \cdot \sin 2\theta d\theta } \\ \end{align}

You can proceed further as we did from (1) above.

Example - 21

Evaluate \begin{align}\int {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}} dx\end{align}

Solution: In our search for an appropriate substitution, we must look into how we can express the numerator $$(\sin x+\cos x)dx$$  in terms of the differential of the denominator (or some part of the denominator).

Notice that $$(\sin x + \cos x)$$ is the derivative of $$( - \cos x + \sin x).$$ What if we could express the $$( - \cos x + \sin x)$$ denominator in terms of . We could then use the substitution $$( - \cos x + \sin x)$$ = y. This is what we proceed to do:

\begin{align}& \quad\;9 + 16\sin 2x\\& = 9 - 16( - 1 + 1 - \sin 2x)\\& = 9 - 16\left\{ { - 1 + {{(\sin x - \cos x)}^2}} \right\}\\& = 25 - 16{( - \cos x + \sin x)^2}\end{align}

Thus, we have succeeded in expressing the denominator in terms of $$( - \cos x + \sin x)$$ :

Substituted $$- \cos x + \sin x = y$$

\begin{align}& \Rightarrow \quad(\sin x + \cos x)dx = dy\\ &\Rightarrow\quad I = \int {\frac{{dy}}{{25 - 16{y^2}}}}\\ &\qquad\quad= \frac{1}{{16}}\int {\frac{{dy}}{{\left( {25{\rm{/}}16} \right) - {y^2}}}}\end{align}

This integral is of the form discussed in Example -18 part (d). You can proceed from here based on the discussion done previously in that example. Verify that the answer is :

$I = \frac{1}{{40}}\ln \left| {\frac{{5 - 4\cos x + 4\sin x}}{{5 + 4\cos x - 4\sin x}}} \right| + C$