Question

\[\int\text{ cos }\left( \text{ log x } \right) \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\text{ cos }\left( \text{ log x } \right) \text{ dx }\]
\[\text{ Let  log x } = t\]
\[ \Rightarrow x = e^t \]
\[ \Rightarrow dx = e^t dt\]
\[I = \int e^t \cos\left( t \right)dt\]
`\text{Considering cos  ( t ) as first function and` `\text{ e}^{t}`   ` \text{ as second function} `
\[I = \text{ cos  t  e}^t - \int \left( - \sin t \right) e^t dt\] 
\[ \Rightarrow I = \text{ cos  t  e}^t + \int \text{ sin t e }^t dt\]
\[ \Rightarrow I = \text{ cos t e}^t + I_1 . . . . . \left( 1 \right)\]
\[\text{ where I}_1 = \int e^t \text{ sin  t  dt }\]
\[ I_1 = \int e^t \text{ sin  t  dt}\]
\[\text{Cosidering sin t as first function and e}^t \text{ as second function}\]
\[ I_1 = \text{ sin t  e}^t - \int \text{ cos t e }^t  \text{ dt }\]
\[ \Rightarrow I_1 = \text{ sin  t  e}^t - I . . . . . \left( 2 \right)\]
` \text{ From ( 1 ) and ( 2) } `
\[I = \text{ cos  t  e}^t + \text{ sin t  e}^t - I\]
\[ \Rightarrow 2I = e^t \left( \sin t + \cos t \right)\]
\[ \Rightarrow I = \frac{e^t \left( \sin t + \cos t \right)}{2} + C\]
\[ \Rightarrow I = \frac{e^\text{ log  x }\left[  sin \left( \log x \right) +  cos\left( \log x \right) \right]}{2} + C\]
\[ \Rightarrow I = \frac{x}{2}\left[ \text{ sin } \left( \log x \right) + \text{ cos } \left( \log x \right) \right] + C\]