Evaluate the following integrals:

Let $I=\int \frac{\cos x}{\sin ^{2} x+4 \sin x+5} d x$

Let $\sin x=t \ldots . .(i)$

$\Rightarrow \cos x d x=d t$

$S O, I=\int \frac{d t}{t^{2}+4 t+5}$

$=\int \frac{d t}{t^{2}+(2 t)(2)+2^{2}-2^{2}+5}$

$\int \frac{d t}{(t+2)^{2}+1}$

Again, let $t+2=u \ldots$ (ii)

$\Rightarrow \mathrm{dt}=\mathrm{du}$

$I=\int \frac{d u}{u^{2}+1}$

$=\tan ^{-1} u+c$

[since, $\left.\int \frac{1}{1+(\mathrm{x})^{2}} \mathrm{dx}=\tan ^{-1} \mathrm{x}+\mathrm{c}\right]$

$=\tan ^{-1}(\sin x+2)+c[$ using $(i),(i i)]$