Question

Evaluate : \[\int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx\] .

Answer 1

\[\text { Let } I = \int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx\]

\[\text { Put} \cos^{- 1} x = t\]

\[\text { Differentiating with respect to } x: \]

\[\frac{- 1}{\sqrt{1 - x^2}}dx = dt\]

\[I = - \int \text { t cost dt }\]

Applying integration by parts:

\[\int f\left( x \right)g\left( x \right)dx = f\left( x \right)\int g\left( x \right)dx - \int\left( f'\left( x \right)\int g\left( x \right)dx \right)dx\]

\[ \Rightarrow I = - \left[ t\int\cos t dt - \int\left( \frac{d\left( t \right)}{dt}\int\cos t dt \right)dt \right]\]

\[ \Rightarrow I = - \left[ t \sin t - \int\sin t dt \right]\]

\[ \Rightarrow I = - t \sin t - \cos t + C\]

\[I = - t\sqrt{1 - \cos^2 t} - \cos t + C\]

\[\text { Substituting the value of t }: \]

\[I = - \cos^{- 1} x\sqrt{1 - \cos^2 \left( \cos^{- 1} x \right)} - \cos\left( \cos^{- 1} x \right) + C\]

\[I = - \left( \sqrt{1 - x^2} \right) \cos^{- 1} x - x + C\]