Evaluate the following integrals:

Let $\mathrm{I}=\int \sin \mathrm{x} \log (\cos \mathrm{x}) \mathrm{dx}$

Put $\cos x=t$

$-\sin x d x=d t$

$I=\int-\log t d t$

Using integration by parts,

$=\int 1 \times-\log t d t$

$=-\left(\log \mathrm{t} \int \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \log \mathrm{t} \int 1 \mathrm{dt}\right)$

$=-\left(\mathrm{t} \log \mathrm{t}-\int \frac{1}{\mathrm{t}} \times \mathrm{t} \mathrm{dt}\right)$

$=-\left(\mathrm{t} \log \mathrm{t}-\int \mathrm{dt}\right)$

$=-(\mathrm{t} \log \mathrm{t}-\mathrm{t})+\mathrm{c}$

$=\mathrm{t}(1-\log \mathrm{t})+\mathrm{c}$

Replace t by $\cos x$

$I=\cos x(1-\log (\cos x))+c$