Evaluate the following integrals:

Assume $2+3 \log x=t$

$\mathrm{d}(2+3 \log \mathrm{x})=\mathrm{dt}$

$\Rightarrow \frac{3}{x} d x=d t$

$\Rightarrow \frac{1}{x} d x=\frac{d t}{3}$

Substituting $t$ and $d t$

$\Rightarrow \frac{1}{3} \int \sin t d t$

$=-\cos t+c$

But $t=2+3 \log x$

$\Rightarrow \frac{-1}{3} \cos (2+3 \log x)+c$