Question

$\int \frac{dx}{3+2sinx+cosx}$

Answer 1

We have,

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

$=\int \frac{dx}{3+2\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}$

$=\int \frac{(1+{\tan }^2\frac{x}{2})dx}{3(1+{\tan }^2\frac{x}{2})+2\left(2\tan \frac{x}{2}\right)+(1-{\tan }^2\frac{x}{2})}$

$=\int \frac{{\sec }^2\frac{x}{2}dx}{3(1+{\tan }^2\frac{x}{2})+2\left(2\tan \frac{x}{2}\right)+(1-{\tan }^2\frac{x}{2})}$

$=\int \frac{{\sec }^2\frac{x}{2}dx}{3+3{\tan }^2\frac{x}{2}+4\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}}$

$=\int \frac{{\sec }^2\frac{x}{2}dx}{4+2{\tan }^2\frac{x}{2}+4\tan \frac{x}{2}}$

$=\frac{1}{2}\int \frac{{\sec }^2\frac{x}{2}dx}{{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}+2}$

$=\frac{1}{2}\int \frac{{\sec }^2\frac{x}{2}dx}{{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}+1+1}$

$=\frac{1}{2}\int \frac{{\sec }^2\frac{x}{2}dx}{{(\tan \frac{x}{2}-1)}^2+1}$

Let

$\tan \frac{x}{2}=t$

$\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$

${\sec }^2\frac{x}{2}dx=2dt$

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

$={\tan }^{-1}(t+1)+C$

$={\tan }^{-1}(\tan \frac{x}{2}+1)+C$

Hence, this is the answer.