Question

Find $\int \frac{1}{3+2\sin x+\cos x}dx$.

Answer 1

Let $\int \frac{dx}{3+2\sin x+\cos x}$

Put $\tan \left(\frac{x}{2}\right)=t$

$\therefore x=2{\tan }^{-1}t$

$\therefore dx=\frac{2dt}{1+t^2}$ and $\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$

$\therefore I=\int \frac{1}{3+2\left(\frac{2t}{1+t^2}\right)+\left(\frac{1-t^2}{1+t^2}\right)}.\frac{2dt}{1+t^2}$

$=\int \frac{1+t^2}{3(1+t^2)+4t+(1-t^2)}.\frac{2dt}{1+t^2}$

$=2\int \frac{dt}{2t^2+4t+4}$

$=\int \frac{dt}{t^2+2t+2}$

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

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

$=\frac{1}{1}{\tan }^{-1}\left(\frac{t+1}{1}\right)+c$

$={\tan }^{-1}[\tan \left(\frac{x}{2}\right)+1]+c$