Evaluate the following integrals:

Assume $2 \sin x+\cos x=t$

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

$(2 \cos x-\sin x) d x=d t$

Put $t$ and dt in given equation we get

$\Rightarrow \int \frac{\mathrm{d} t}{t}$

$=\ln |t|+c$

But $t=2 \sin x+\cos x$

$=\ln |2 \sin x+\cos x|+c$