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Question

Mathematics Question on integral

Integrate the function: (sin−1x)2(sin^{-1}x)^2

Answer

The correct answer is: =x(sin−1x)2+21−x2sin−1x−2x+C=x(sin^{-1}x)^2+2\sqrt{1-x^2}sin^{-1}x-2x+C
Let I=∫(Sin−1x)2.1dxI=∫(Sin^{-1}x)^2.1 dx
Taking(sin−1x)2(sin^{-1}x)^2 as first function and 1 as second function and integrating by parts,we
obtain
I=(sin−1x)∫1dx−∫[ddx(sin−1x)2.∫1.dx]dxI=(sin^{-1}x)∫1dx-∫[{\frac{d}{dx}(sin^{-1}x)^2.∫1.dx}]dx
=(sin−1x)2.x−∫2sin−1x1−x2.xdx=(sin^{-1}x)^2.x-∫\frac{2sin^{-1}x}{\sqrt{1-x^2}}.x dx
=x(sin−1x)2+∫sin−1x.(−2x1−x2)dx=x(sin^{-1}x)^2+∫sin^{-1}x.(\frac{-2x}{\sqrt{1-x^2}})dx
=x(sin−1x)2+[sin−1x∫−2x1−x2dx−∫(ddxsin−1x)∫−2x1−x2dx]dx=x(sin^{-1}x)^2+[sin^{-1}x∫\frac{-2x}{\sqrt{1-x^2}} dx-∫{(\frac{d}{dx} sin^{-1}x)∫\frac{-2x}{\sqrt{1-x^2}} dx}]dx
=x(sin−1x)2+[sin−1x.21−x2−∫11−x2.21−x2dx]=x(sin^{-1}x)^2+[sin^{-1}x.2\sqrt{1-x^2}-∫\frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2} dx]
=x(sin−1x)2+21−x2sin−1x−∫2.dx=x(sin^{-1}x)^2+2\sqrt{1-x^2} sin^{-1}x-∫2.dx
=x(sin−1x)2+21−x2sin−1x−2x+C=x(sin^{-1}x)^2+2\sqrt{1-x^2}sin^{-1}x-2x+C