Question

\[\int\frac{1}{x^3}\text{ sin } \left( \text{ log x }\right) dx\]

Answer 1

\[\text{ Let I } = \int \frac{1}{x^3}\text{ sin} \left( \text{ log x } \right)\text{ dx }\]
\[\text{ Putting log x }= t\]
\[ \Rightarrow x = e^t \]
\[ \Rightarrow dx = e^t dt\]
\[ \therefore I = \int \frac{1}{e^{3t}}\text{ sin t e}^t dt\]
\[ = \int e^{- 2t} \text{ sin  t  dt }\]
`\text{Considering    sin  t  as first function and` `\text{ e}^{-2t}`   ` \text{ as second function} `
\[I = \sin t\left[ \frac{e^{- 2t}}{- 2} \right] - \int \cos t\frac{e^{- 2t}}{- 2}dt\]
\[ \Rightarrow I = \frac{\text{ sin  t  e}^{- 2t}}{- 2} + \frac{1}{2}\int\cos t e^{- 2t} dt\]
\[ \Rightarrow I = \frac{\text{ sin  t  e}^{- 2t}}{- 2} + \frac{1}{2}\left[ \cos t\frac{e^{- 2t}}{- 2} - \int\left( - \sin t \right)\frac{e^{- 2t}}{- 2}dt \right]\]
\[ \Rightarrow I = \frac{\text{ sin  t  e}^{- 2t}}{- 2} - \frac{1}{4} \text{ cos  t  e}^{- 2t} - \int \frac{e^{- 2t} \text{ sin  t  dt}}{4}\]
\[ \Rightarrow I = e^{- 2t} \left[ \frac{- 2 \sin t - \cos t}{4} \right] - \frac{I}{4}\]
\[ \Rightarrow \frac{5I}{4} = e^{- 2t} \left[ \frac{- 2 \sin t - \cos t}{4} \right]\]
\[ \Rightarrow I = \frac{e^{- 2t}}{5}\left[ - 2 \sin t - \cos t \right] + C\]
\[ \Rightarrow I = \frac{- x^{- 2}}{5}\left[ 2 \text{ sin }\left( \log x \right) + \text{ cos }\left( \log x \right) \right] + C\]
\[ \Rightarrow I = \frac{- 1}{5 x^2}\left[ \text{ cos }\left( \log x \right) + 2 \text{ sin }\left( \log x \right) \right] + C\]