#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 15 Maths Textbook Solution.

Hint: Use indefinite formula then put the limit to get the required answer

Given:$\int_{\pi }^{0}\frac{1}{1+\sin x}dx$

Solution:$\int_{\pi }^{0}\frac{1}{1+\sin x}dx$

Rationalizing,$\int_{0}^{\pi}\left(\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}\right) d x=\int_{0}^{\pi}\left(\frac{1-\sin x}{(1+\sin x)(1-\sin x)}\right)$

\begin{aligned} &=\int_{0}^{\pi} \frac{1-\sin x}{1-\sin ^{2} x} d x \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right] \\ &=\int_{0}^{\pi} \frac{1-\sin x}{\cos ^{2} x} d x \quad\left[1-\sin ^{2} x=\cos ^{2} x\right] \end{aligned}

\begin{aligned} &=\int_{0}^{\pi}\left(\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right) d x \\ &=\int_{0}^{\pi}\left(\sec ^{2} x-\frac{\sin x}{\cos x} \frac{1}{\cos x}\right) d x \end{aligned}

\begin{aligned} &=\int_{0}^{\pi}\left(\sec ^{2} x-\tan x \sec x\right) d x \quad\left[\frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\right] \\ &=\int_{0}^{\pi} \sec ^{2} x-\int_{0}^{\pi} \tan x \sec x d x \quad\left[\begin{array}{l} \int \sec ^{2} x d x=\tan x \\ \int \sec x \tan x d x=\sec x \end{array}\right] \end{aligned}

$\left.\begin{array}{l} =[\tan x]_{0}^{\pi}-[\sec x]_{0}^{\pi} \\ =[\tan \pi-\tan 0]-[\sec \pi-\sec 0] \end{array}\right]\left[\begin{array}{l} \tan \pi=\tan 0=0 \\ \sec \pi=-1, \sec 0=1 \end{array}\right]$

$=\left [ 0-0 \right ]-\left [ -1-1 \right ]$

$=0-\left ( -2 \right )$

$=2$