Question

` ∫    sin x log  (\text{ cos x ) } dx  `

Answer 1

 ` \text{ Let I }= ∫    sin x  . log  (\text{ cos x ) } dx  `
\[\text{ Let cos x }= t\]
\[ \Rightarrow - \text{ sin x dx }= dt\]
\[ \Rightarrow \text{ sin x dx }= - dt \]
\[ \therefore I = - \int\text{ log t dt}\]
\[ = - \int 1_{} \cdot \text{ log t dt }\]
\[\text{Taking log t as the first function and 1 as the second function} . \]
\[ = \log t\int \text{ 1 dt }- \int\left\{ \frac{d}{dt}\left( \log t \right)\int1dt \right\}dt\]
\[ = - \left[ \log t \cdot t - \int\frac{1}{t} \times\text{  t dt } \right]\]
\[ = - \left[ \log t \cdot t - t \right] + C\]
\[ = - t\left( \log t - 1 \right) + C . . . . (1) \]
\[\text{Substituting the value of t in eq}   \text{ (1) }\]
\[ = - \cos x\left\{ \text{ log  }\left( \text{ cos x }\right) - 1 \right\} + C\]
\[ = \text{ cos x }\left\{ 1 - \text{ log }\left( \cos x \right) \right\} + C\]