Evaluate the following integrals:

Assume $2+3 \sin x=t$

$d(2+3 \sin x)=d t$

$3 \cos x d x=d t$

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

Put $t$ and dt in given equation we get

$\Rightarrow \frac{1}{3} \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\frac{1}{3} \ln |\mathrm{t}|+\mathrm{c}$

But $t=2+3 \sin x$

$=\frac{1}{3} \ln |2+3 \sin x|+c$