Mathematics

# $\int {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \,dx =$

##### SOLUTION

Put $x=\cos 2\theta$    …….  (1)

On differentiation with respect to $\theta$ , and we get

$dx=-2\sin 2\theta d\theta$

put given function  and solve given function

$\int{\sqrt{\dfrac{1-x}{1+x}}}dx$

$=\int{\sqrt{\dfrac{1-\cos 2\theta }{1+\cos 2\theta }}\left( -2\sin 2\theta d\theta \right)}$

$=-2\int{\sqrt{\dfrac{1-\left( 1-2{{\sin }^{2}}\theta \right)}{1+\left( 2{{\cos }^{2}}\theta -1 \right)}}\sin 2\theta d\theta }$

$=-2\int{\sqrt{\dfrac{2{{\sin }^{2}}\theta }{2{{\cos }^{2}}\theta }}}\sin 2\theta d\theta$

$=-2\int{\sqrt{{{\tan }^{2}}\theta }}\sin 2\theta d\theta$

$=-2\int{\tan \theta }\sin 2\theta d\theta$

$=-2\int{\dfrac{\sin \theta }{\cos \theta }2\sin \theta \cos \theta d\theta }$

$=-4\int{{{\sin }^{2}}\theta d\theta }$

$=-4\int{\left( \dfrac{1-\cos 2\theta }{2} \right)}d\theta$

$=-2\int{1d\theta +2\int{\cos 2\theta d\theta }}$

On integration, we get

$=-2\theta +2\dfrac{\sin 2\theta }{2}+C$

$=-2\theta +\sin 2\theta +C$

By equation (1),

$\theta =\dfrac{1}{2}{{\cos }^{-1}}x$

Put here and we get,

$=-2\times \dfrac{1}{2}{{\cos }^{-1}}x+\sin 2\left( \dfrac{1}{2}{{\cos }^{-1}}x \right)+C$

$=-{{\cos }^{-1}}x+\sin {{\cos }^{-1}}x+C$

$=-{{\cos }^{-1}}x+\sin {{\sin }^{-1}}\sqrt{1-{{x}^{2}}}+C$

$=-{{\cos }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$

$=\sqrt{1-{{x}^{2}}}-{{\cos }^{-1}}x+C$

It is complete solution.

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

#### Realted Questions

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If $f\left( x \right)$ is function satisfying $f\left( {\dfrac{1}{x}} \right) + {x^2}f\left( x \right) = 0$ for all non-zero x, then $\int\limits_{\sin \theta }^{\cos ec\theta } {f\left( x \right)dx}$ equals
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Q4 One Word Hard
*$\displaystyle\int \sqrt{\left ( \frac{x}{a^{3}-x^{3}} \right )\cdot }dx=\frac{6}{k}\sin ^{-1}\frac{x^{3/2}}{a^{3/2}}.$ Find the value of $k$.

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