#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 16 Maths Textbook Solution.

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

Given: $\int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin x}dx$

Solution:

$\int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin x}dx$

\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}\right) d x \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1-\sin x}{1-\sin ^{2} x}\right) d x \end{aligned}$\left [ \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2}\right ]$

$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1-\sin x}{\cos ^{2} x}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1-\sin ^{2} x=\cos ^{2} x\right]$

\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right) d x \text {. } \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sec ^{2} x-\frac{\sin x}{\cos x} \frac{1}{\cos x}\right) d x \end{aligned} \quad\left[\begin{array}{l} \frac{1}{\cos x}=\sec x \\ \frac{\sin x}{\cos x}=\tan x \end{array}\right]

\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sec ^{2} x-\tan x \sec x\right) d x \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec x \tan x d x \end{aligned}                                                                    $\left[\begin{array}{l} \int \sec x \tan x d x=\sec x \\ \int \sec ^{2} x d x=\tan x \end{array}\right]$

$=[\tan x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}-[\sec x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$

\begin{aligned} &=\left[\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)\right]-\left[\sec \frac{\pi}{4}-\sec \left(\frac{-\pi}{4}\right)\right] \\ &=\left[\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}\right)\right]-\left[\sec \frac{\pi}{4}-\sec \left(\frac{\pi}{4}\right)\right] \end{aligned} \quad\left[\begin{array}{l} \tan (-\theta)=-\tan \theta] \\ \sec (-\theta)=\sec \theta \end{array}\right]

$=\left[\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{4}=1\right]$

$=2\tan \frac{\pi }{4}$

$=2.1$

$=2$