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 Q11  Integrate the functions \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}

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Consider \int \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}dx =I

So, we have then: I = \frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}. \cos^{-1}x dx

After taking \cos ^{-1}x as a first function and \left ( \frac{-2x}{\sqrt{1-x^2}} \right ) as second function and integrating by parts, we get

I =-\frac{1}{2}\left [ \cos^{-1}x\int\frac{-2x}{\sqrt{1-x^2}}dx - \int\left \{ \left ( \frac{d}{dx}\cos^{-1}x \right )\int \frac{-2x}{\sqrt{1-x^2}}dx \right \}dx \right ]=-\frac{1}{2}\left [ \cos^{-1}x.2{\sqrt{1-x^2}} + \int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]

=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-\int2dx \right ]

=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-2x \right ]+C

Or, -\left \sqrt{1-x^2}\cos^{-1}x +x\right +C

Posted by

Divya Prakash Singh

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