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Integrate the function
$\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}$
$\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}$
Solution:
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Verified Answer
Put $\cos ^{-1} x=t$ so that $\frac{-1}{\sqrt{1-x^2}} d x=d t$
$\begin{aligned}
&\therefore \mathrm{I}=-\int t \cos t d t=-\left[t(\sin t)-\int 1 \cdot(\sin t) d t\right. \\
&=-t \sin t-\cos t+\mathrm{C}=-t \sqrt{1-\cos ^2 t}-\cos t+\mathrm{C} \\
&=-\left[\cos ^{-1} x \cdot \sqrt{1-x^2}+x\right]+\mathrm{C}
\end{aligned}$
$\begin{aligned}
&\therefore \mathrm{I}=-\int t \cos t d t=-\left[t(\sin t)-\int 1 \cdot(\sin t) d t\right. \\
&=-t \sin t-\cos t+\mathrm{C}=-t \sqrt{1-\cos ^2 t}-\cos t+\mathrm{C} \\
&=-\left[\cos ^{-1} x \cdot \sqrt{1-x^2}+x\right]+\mathrm{C}
\end{aligned}$
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