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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 58

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Answer: -\frac{1}{3} \cos (2+3 \log x)+c

Hint: Use substitution method to solve this integral

Given: \int \frac{\sin (2+3 \log x)}{x} d x


        \text { Let } I=\int \frac{\sin (2+3 \log x)}{x} d x

        \begin{aligned} &\text { Put } 2+3 \log x=t \Rightarrow 3 \cdot \frac{1}{x} d x=d t \Rightarrow d x=\frac{x}{3} d t \\ &\text { Then, } I=\int \frac{\sin t}{x} \cdot \frac{x}{3} d t=\frac{1}{3} \int \sin t \; d t \end{aligned}                

                        =\frac{1}{3}(-\cos t)+c \quad\left[\because \int \sin x\; d x=-\cos x+c\right]

                        \begin{aligned} &=\frac{-1}{3} \cos t+c \\ &=\frac{-1}{3} \cos (2+3 \log x)+c \end{aligned}            [\because t=2+3 \log x]

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