#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 42 Maths Textbook Solution.

Answer: $2\: x\tan ^{-1}x-2log\left ( \sqrt{1+x^{2}} \right )+c$

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

Hint: Put $x=\tan \Theta$

Solution:

$I=\int \cos ^{-1}\left ( \frac{1-x^{2}}{1+x^{2}} \right )dx$

Put $x=\tan \theta\Rightarrow dx=\sec ^{2}\theta d\theta$

\begin{aligned} &I=\int \cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta \\ &{\left[\cos 2 \theta=\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\right]} \\ &I=\int \cos ^{-1}(\cos 2 \theta) \sec ^{2} \theta d \theta \end{aligned}

\begin{aligned} &=\int 2 \theta \sec ^{2} \theta d \theta \\ &=2 \int \theta \sec ^{2} \theta d \theta \\ &=2\left[\theta \int \sec ^{2} \theta d \theta-\int \frac{d}{d x} \theta\left(\int \sec ^{2} \theta d \theta\right) d \theta\right] \\ &=2 \theta \times \tan \theta-2 \int \tan \theta d \theta \\ &=2 \theta \tan \theta-2 \theta+c \\ &\theta=\tan ^{-1} x \\ &=2 x \tan ^{-1} x-2 \log \left(\sqrt{1+x^{2}}\right)+c \\ &=2 x \tan ^{-1} x-2 \log \left(\sqrt{1+x^{2}}\right)+c \end{aligned}