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

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

Answers (1)

Answer: 2

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]

 

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