Question
Integrate the function: $\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$

Answer

Let $I=\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$ 
$I=-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$ 
Now, integrating by parts, we get,
$I = \frac{-1}{2}\left[\cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\right\} d x\right]$ 
= $-\frac{1}{2}\left[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x\right]$ 
= $-\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x\right]$ 
= $-\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x\right]+C$ 
= $-\left[\sqrt{1-x^{2}} \cos ^{-1} x+x\right]+C$ 

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