Question

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

Answer 1

The correct answer is: \(=x(sin^{-1}x)^2+2\sqrt{1-x^2}sin^{-1}x-2x+C\)
Let \(I=∫(Sin^{-1}x)^2.1 dx\)
Taking\((sin^{-1}x)^2\) as first function and 1 as second function and integrating by parts,we
obtain
\(I=(sin^{-1}x)∫1dx-∫[{\frac{d}{dx}(sin^{-1}x)^2.∫1.dx}]dx\)
\(=(sin^{-1}x)^2.x-∫\frac{2sin^{-1}x}{\sqrt{1-x^2}}.x dx\)
\(=x(sin^{-1}x)^2+∫sin^{-1}x.(\frac{-2x}{\sqrt{1-x^2}})dx\)
\(=x(sin^{-1}x)^2+[sin^{-1}x∫\frac{-2x}{\sqrt{1-x^2}} dx-∫{(\frac{d}{dx} sin^{-1}x)∫\frac{-2x}{\sqrt{1-x^2}} dx}]dx\)
\(=x(sin^{-1}x)^2+[sin^{-1}x.2\sqrt{1-x^2}-∫\frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2} dx]\)
\(=x(sin^{-1}x)^2+2\sqrt{1-x^2} sin^{-1}x-∫2.dx\)
\(=x(sin^{-1}x)^2+2\sqrt{1-x^2}sin^{-1}x-2x+C\)