Question

Integrate the function: \(\frac{xcos^{-1}x}{\sqrt{1-x^2}}\)

Answer 1

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