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Find : \int \sin ^-^1(2x)\: dx

 

 

 

 
 
 
 
 

Answers (1)

Let      I=\int \sin^{-1}2x\: dx

\sin^{-1}2x=t

2x=\sin t

Differentiating with respect to x,

2\: dx=\cos t\: dt

dx=\frac{\cos t\: dt}{2}

Substituting in I,

I=\frac{1}{2}\int t\cos t \: dt

=\frac{1}{2}\left [ t\: \sin t-\int 1\cdot \sin t\: dt \right ]     (using by parts)

=\frac{1}{2}\left [ t\: \sin t-\cos t \right ]+c

\left [ Substituting \: \: t=sin^{-1}2x\: \: and \: \: 2x=\sin t \right ]

=\frac{1}{2}\left [ 2x\: \sin^{-1} 2x-\cos (\sin^{-1}2x) \right ]+c

=\frac{1}{2}\left [ 2x\: \sin^{-1} 2x-\sqrt{1-4x^2} \right ]+c

Posted by

Ravindra Pindel

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