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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 9

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Answer: I=-\frac{1}{5x}\left [ 2\sin \left ( \log x \right ) +\cos \left ( \log x \right )\right ]+c


Let t=log x\\dt=\frac{dx}{e^{\log x}}

Given: \int \frac{1}{x^{3}}\sin \left ( \log x \right )dx

Solution: I=\int e^{-4t}\sin t dt

\begin{aligned} &{\left[e^{a x} \sin b x d x=\frac{e^{a x}}{a^{2}+b^{2}}\{a \sin b x-b \cos b x\}+c\right]} \\ &=\frac{e^{-4 t}}{17}(-4 \sin t-\cos t)+c \\ &=\frac{e^{-4t}}{17}(-4 \sin t+\cos t)+c \\ &=\frac{e^{-4 \log x}}{17}[-4 \sin (\log x)+\cos (\log x)]+c \\ &=-\frac{1}{17 x^{4}}[-4 \sin (\log x)+\cos (\log x)]+c \end{aligned}


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