Find: \int \ \sin x\cdot \log\cos x\ \text d x.

 

 

 

 
 
 
 
 

Answers (1)

I = \int \ \underline{\sin x}\cdot \log\cos x\ \underline{\text d x}

Put 

\\\cos x = t\\ \sin x dx = - dt

I = -\int \log t \ \text d t \Rightarrow -\int \underset{\text{I}}{\log t} \underset{\text{II}}{\text d t}

I =- \left \{ \log t \cdot t- \int \frac{1}{t}t\ d t\right\}    [Using by parts]

I = -t\log t + t +c

    = -\cos x\left (\log\cos x -1 \right ) +c

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