Question

Evaluate the following integrals:

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

 

Answer 1

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

\[\]

\[\text{ Let the first function be} \cos^{- 1} \text{ x and second function be} \frac{x}{\sqrt{1 - x^2}} . \]

\[\text{ First we find the integral of the second function}, i . e . , \int\frac{x}{\sqrt{1 - x^2}}dx . \]

\[\text{ Put t } = 1 - x^2 . Then dt = - 2xdx\]

\[\]

\[\text{ Therefore,} \]

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

\[ = - \sqrt{t}\]

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

\[\]

\[\text{ Hence, using integration by parts, we get }\]

\[\int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx = \left( \cos^{- 1} x \right)\int\frac{x}{\sqrt{1 - x^2}}dx - \int\left[ \left( \frac{d \left( \cos^{- 1} x \right)}{d x} \right)\int\left( \frac{x}{\sqrt{1 - x^2}}dx \right) \right]dx\]

\[ = \left( \cos^{- 1} x \right)\left( - \sqrt{1 - x^2} \right) - \int\left( \frac{- 1}{\sqrt{1 - x^2}} \right)\left( - \sqrt{1 - x^2} \right)dx\]

\[ = - \sqrt{1 - x^2} \cos^{- 1} x - x + c\]

\[\]

\[\]

\[\text{ Hence}, \int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx = - \sqrt{1 - x^2} \cos^{- 1} x - x + c\]