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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 3

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Answer: I=\frac{x}{2}\left [ \sin \left ( \log x \right ) +\cos \left ( \log x \right )\right ]+c

Hint: Using ILATE rule

Given: \int \cos \left ( \log x \right )dx


\begin{aligned} &\text { Let } I=\int \cos (\log x) d x\\ &\text { Integrating by parts }\\ &u=\cos (\log x)\\ &d u=-\sin (\log x) \frac{d x}{x}\\ &\\ &\int \cos (\log x) d x=x \cos (\log x)+\int \sin (\log x) d x \end{aligned}

\begin{aligned} &\text { Again Integrating by parts }\\ &u=\sin (\log x)\\ &d u=\cos (\log x) \frac{d x}{x}\\ &\\ &\int \cos (\log x) d x=x \cos (\log x)+x \sin (\log x)-\int \cos (\log x) d x \end{aligned}

\begin{aligned} &I=2 \int \cos (\log x) d x=x \cos (\log x)+x \sin (\log x) \\ &I=\int \cos (\log x) d x=\left(\frac{1}{2}\right) \times[\cos (\log x)+\sin (\log x)]+c \\ &I=\frac{1}{2} x[\cos (\log x)+\sin (\log x)]+c \end{aligned}

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