Question:
Evaluate the following integrals:
$\int \frac{\cos x}{\sin ^{2} x+4 \sin x+5} d x$
Solution:
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 $(i),(i i)]$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.