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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18 .9 question 42

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

Hint: Use substitution method to solve this integral.

Given:   \int \log x \cdot \frac{\sin \left\{1+(\log x)^{2}\right\}}{x} d x


        \begin{aligned} &\text { Let } I=\int \log x \cdot \frac{\sin \left\{1+(\log x)^{2}\right\}}{x} d x \\ &\text { Put } 1+(\log x)^{2}=t \Rightarrow 2 \log x \cdot \frac{1}{x} \cdot d x=d t \\ &\Rightarrow d x=\frac{x \cdot d t}{2 \log x} \text { then } \end{aligned}

        \begin{aligned} \Rightarrow I &=\int \log x \cdot \frac{\sin t}{x} \cdot \frac{x \; d t}{2 \log x} d t \\ &=\frac{1}{2} \int \sin t\; d t \end{aligned}

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

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