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Q.
$\int \sin 2 x .\log \cos x d x$ is equal to
Integrals
Solution:
$I =\int 2 \sin x \cdot \cos x \cdot \log \cos x dx$
put $\log \cos x=t$
$\therefore -\frac{\sin x}{\cos x} d x=d t$
$I=\int 2 \sin x \cdot \cos x \cdot t \frac{\cos x}{-\sin x} d t$
$=-2 \int \cos ^{2} x . t dt =-2 \int te ^{2 t} dt$
$=-2\left[ t \cdot \frac{ e ^{2 t }}{2}-\int \frac{ e ^{2 t }}{2} \cdot dt \right]=- t e ^{2 t }+\frac{1}{2} e ^{2 t }+ k$
$= e ^{2 t }\left(\frac{1}{2}- t \right)+ k =\cos ^{2} x \left\{\frac{1}{2}-\log \cos x \right\}+ k$