#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.13 Question 3

Answer: $\frac{x^{2}}{2}+c$

Hint: Use substitution method to solve this integral

Given: $\int \cos \left \{ 2\cot^{-1} \sqrt{\frac{1+x}{1-x}}\right \}dx$

Put$x=\cos2\theta$                                                             …(i)

\begin{aligned} &\text { Then } I=\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}\right\} d x \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{2 \cos ^{2} \theta}{2 \sin ^{2} \theta}}\right\} d x \quad\quad\quad\quad\quad\left\{\begin{array}{l} \because 1+\cos 2 A=2 \cos ^{2} A \\ 1-\cos 2 A=2 \sin ^{2} A \end{array}\right\} \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{\left(\frac{\cos \theta}{\sin \theta}\right)^{2}}\right\} d x \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{(\cot \theta)^{2}}\right\} d x \quad\quad\quad\quad\quad\quad\left[\because \frac{\cos A}{\sin A}=\cot A\right] \\ &=\int \cos \left\{2 \cot ^{-1} \cdot \cot \theta\right\} d x \end{aligned}

\begin{aligned} &=\int \cos \{2 \theta\} d x \quad\left[\because \cot \left(\cot ^{-1} x\right)=x\right] \\ &=\int \cos \left\{\cos ^{-1}(x)\right\} d x \\ &=\int x^{1} d x \quad\left[\begin{array}{c} \because x=\cos 2 \theta \\ 2 \theta=\cos ^{-1}(x) \end{array}\right] \\ &=\frac{x^{1+1}}{1+1}+c \quad\left[\because \cos \left(\cos ^{-1} x\right)=x\right] \\ &=\frac{x^{2}}{2}+c \end{aligned}