Question:

Integrate the function: tan11x1+xtan^{-1}\sqrt{\frac{1-x}{1+x}}

Updated On: Oct 7, 2023
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Solution and Explanation

I=tan11x1+xI=tan^{-1}\sqrt{\frac{1-x}{1+x}}

Let x=cosθdx=sinθdθLet\space x=cosθ⇒dx=-sinθdθ

I=tan11cosθ1+cosθ(sinθdθ)I=\int tan^{-1}\sqrt{\frac{1-cosθ}{1+cosθ}}(-sinθdθ)

=tan12sin2θ22cos2θ2sinθdθ=-\int tan^{-1}\sqrt{\frac{2sin^{2}\frac{θ}{2}}{2cos^{2}\frac{θ}{2}}}sinθdθ

=tan1tanθ2.sinθdθ=-\int tan^{-1}tan\frac{θ}{2}.sinθdθ

=12θ.sinθdθ=-\frac{1}{2}∫θ.sinθdθ

=12[θ.(cosθ)1.(cosθ)dθ]=-\frac{1}{2}[θ.(-cosθ)-∫1.(-cosθ)dθ]

=12[θcosθ+sinθ]=-\frac{1}{2}[-θcosθ+sinθ]

=+12θcosθ12sinθ=+\frac{1}{2}θcosθ-\frac{1}{2}sinθ

=12cos1x.x121x2+C=\frac{1}{2}cos^{-1}x.x-\frac{1}{2}\sqrt{1-x^{2}}+C

=x2cos1x121x2+C=\frac{x}{2}cos^{-1}x-\frac{1}{2}\sqrt{1-x^{2}}+C

=12(xcos1x1x2)+C=\frac{1}{2}(xcos^{-1}x-\sqrt{1-x^{2}})+C

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