Mathematics

# $\displaystyle \int \sqrt{\dfrac{a+x}{a-x}}dx$ is equal to-

$a\sin^{-1} (x/a)-\sqrt{a^{2}-x^{2}}+c$

##### SOLUTION

Consider the given integral.

$I=\int{\sqrt{\dfrac{a+x}{a-x}}dx}$

$I=\int{\sqrt{\dfrac{a+x}{a-x}}\times \sqrt{\dfrac{a+x}{a+x}}dx}$

$I=\int{\dfrac{a+x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}$

$I=a\int{\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}-\dfrac{1}{2}\int{\dfrac{-2x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}$

$I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\dfrac{1}{2}\int{\left\{ \dfrac{d}{dx}\left( {{a}^{2}}-{{x}^{2}} \right){{\left( {{a}^{2}}-{{x}^{2}} \right)}^{-1/2}} \right\}dx}$

$I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\dfrac{1}{2}\dfrac{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{1/2}}}{\left( 1/2 \right)}+C$

$I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\sqrt{{{a}^{2}}-{{x}^{2}}}+C$

Hence, this is the correct answer.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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