Question

Evaluate the following integrals:

\[\int\cos\left\{ 2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}} \right\}dx\]

Answer 1

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

\[ \text{Let x} = \cos2\theta\]

\[ \text{On differentiating both sides, we get}\

`dx = - 2 sin2θ      d θ  `

`∴ I = -∫ cos { 2 cot^{- 1} \sqrt{{1 + cos 2θ  }/{1 - \cos2 θ }}}  2 sin2θ   d θ  `

`  = -  2 ∫ cos { 2 cot^{- 1} \sqrt{{2cos ^2θ  }/{2  \sin^2 θ }}}  2 sin^2θ   d θ  `

` - 2      ∫  cos { 2 cot^{- 1} (cot θ )}  sin2θ   d θ  `

` - 2      ∫  cos 2θ   sin2θ   d θ  `

` - 2      ∫     sin4θ   d θ  `

\[ = \frac{\cos4\theta}{4} + c_1 \]

\[ = \frac{1}{4}\left( 2 \cos^2 2\theta - 1 \right) + c_1 \]

\[ = \frac{1}{2} x^2 - \frac{1}{4} + c_1 \]

\[ = \frac{1}{2} x^2 + c, \text{where c} = - \frac{1}{4} + c_1 \]

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