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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 29 maths textbook solution

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Answer:\frac{1}{1+\cos x}+C

Hint:Use substitution method to solve this integral.

Given:   \int \frac{\sin x}{(1+\cos x)^{2}} d x


        \begin{aligned} &\text { Let } I=\int \frac{\sin x}{(1+\cos x)^{2}} d x \\ &\operatorname{Put} 1+\cos x=t \Rightarrow-\sin x d x=d t, \text { then } \end{aligned}

                I=\int \frac{1}{t^{2}}(-d t)=-\int \frac{1}{t^{2}} d t=-\int t^{-2} d t

                    =-\left[\frac{t^{-2+1}}{-2+1}\right]+c=-\frac{-t^{-1}}{-1}+c            \left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

                    =\frac{1}{t}+c=\frac{1}{1+\cos x}+c                            [\because t=1+\cos x]

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