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Question

Mathematics Question on integral

Integrate the function: xcos−1x1−x2\frac{xcos^{-1}x}{\sqrt{1-x^2}}

Answer

The correct answer is: =−[1−x2cos−1x+x]+C=-[\sqrt{1-x^2}cos^{-1}x+x]+C
Let I=∫xcos−1x1−x2dxI=∫\frac{xcos^{-1}x}{\sqrt{1-x^2}} dx
I=−12∫−2x1−x2.cos−1xdxI=\frac{-1}{2} ∫\frac{-2x}{\sqrt{1-x^2}}.cos^{-1}xdx
Taking cos−1xcos^{-1}x as first function and(−2x1−x2)(\frac{-2x}{\sqrt{1-x^2}})as second function and integrating by parts,
we obtain
I=−12[cos−1x∫−2x1−x2dx−∫(ddxcos−1x)∫−2x1−x2dx]dx]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]
=−12[cos−1x.21−x2−∫−11−x2.21−x2dx]=\frac{-1}{2}[cos^{-1}x.2\sqrt{1-x^2}-∫\frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2} dx]
=−12[21−x2 cos−1x+∫2dx]=\frac{-1}{2}[2\sqrt{1-x^2}\,cos^{-1}x+∫2 dx]
=−12[21−x2cos−1x+2x]+C=\frac{-1}{2}[2\sqrt{1-x^2}cos^{-1}x+2x]+C
=−[1−x2cos−1x+x]+C=-[\sqrt{1-x^2}cos^{-1}x+x]+C