Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 10

Hint:  Use trigonometric values, $\int \sec ^{2} x\; {\mathrm{d} \mathrm{x}} \text { and } \int \sec x \tan x \; \mathrm{dx}$

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

Solution:

\begin{aligned} &\mathrm{I}=\int_{0}^{\pi} \frac{1}{1+\sin x} d x \\\\ &\frac{1}{1+\sin x}=\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{1-\sin ^{2} x}=\frac{1-\sin x}{\cos ^{2} x} \end{aligned}

$=\int_{0}^{\pi} \sec ^{2} x\; {d x}-\int_{0}^{\pi} \tan x \sec x$                $\left[\because \frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\right]$

\begin{aligned} &=[\tan x]_{0}^{\pi}-[\sec x]_{0}^{\pi} \\\\ &=\tan (\pi)-\tan (0)-\sec (\pi)+\sec (0) \end{aligned}

\begin{aligned} &=0-0-(-1)+1 \\\\ &=1+1 \\\\ &=2 \end{aligned}                                        $\left[\begin{array}{c} \tan (\pi)=0 ; \tan (0)=0 \\\\ \sec (\pi)=-1 ; \sec (0)=1 \end{array}\right]$