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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 27

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Answer: \frac{1}{2 b(a+b \cos 2 x)}+c

Hint:Use substitution method to solve this integral.

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


        \text { Let } I=\int \frac{\sin 2 x}{(a+b \cos 2 x)^{2}} d x

        \begin{aligned} &\text { Put } a+b \cos 2 x=t \\ &\Rightarrow b(-\sin 2 x) \cdot 2 d x=d t \\ &\Rightarrow-2 b \sin 2 x \; d x=d t \\ &\Rightarrow \sin 2 x \; d x=\frac{d t}{-2 b} \end{aligned}

        I=\int \frac{1}{t^{2}} \cdot \frac{d t}{-2 b}=\frac{-1}{2 b} \int \frac{1}{t^{2}} d t                    

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

            =\frac{-1}{2 b} \cdot \frac{t^{-1}}{-1}+c=\frac{1}{2 b} \cdot \frac{1}{t}+c

            =\frac{1}{2 b(a+b \cos 2 x)}+c \quad \quad[\because t=a+b \cos 2 x]

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