Question

$\int \frac{1}{2+\sin x+\cos x}dx=\sqrt{k}{\tan }^{-1}\frac{[\tan \left(\frac{x}{2}\right)+1]}{\sqrt{\left(k\right)}}.$ Find the value of $k$.

Answer 1

Let $I=\int \frac{1}{2+\sin x+\cos x}dx$
$=\int \frac{1}{2+2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)+{\cos }^2\left(\frac{x}{2}\right)-{\sin }^2\left(\frac{x}{2}\right)}dx$
Divide numerator and denominator by ${\cos }^2\left(\frac{x}{2}\right)$
$I=\int \frac{{\sec }^2\left(\frac{x}{2}\right)dx}{2[1+{\tan }^2\left(\frac{x}{2}\right)]+2\tan \left(\frac{x}{2}\right)+1-{\tan }^2\left(\frac{x}{2}\right)}$

Put $\tan \frac{x}{2}=t$, we get
$I=\int \frac{2dt}{t^2+2t+3}=2\int \frac{dt}{{(t+1)}^2+{\left(\sqrt{2}\right)}^2}$
$=2\frac{1}{\sqrt{\left(2\right)}}{\tan }^{-1}\frac{t+1}{\sqrt{\left(2\right)}}=\sqrt{2}{\tan }^{-1}\frac{[\tan \left(\frac{x}{2}\right)+1]}{\sqrt{\left(2\right)}}$

Hence, $k=2$.