Question

Solve ${\int }_0^{\pi }\frac{1}{3+2\sin x+\cos x}dx$

• A. $\frac{5\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $-\frac{\pi }{4}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{4}$

Let $I_1={\int }_0^{\pi }\frac{1}{3+2\sin x+\cos x}$

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

$at=\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx$

$\Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2dt$

$=\int \frac{1}{\frac{2.2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\frac{1+{\tan }^2\left(\frac{x}{2}\right)}{1-{\tan }^2\left(\frac{x}{2}\right)}+3}dx$

$=\int \frac{1+{\tan }^2\left(\frac{x}{2}\right)}{4\tan \left(\frac{x}{2}\right)+1-{\tan }^2\left(\frac{x}{2}\right)+3+3{\tan }^2\left(\frac{x}{2}\right)}dx$

$=\int \frac{{\sec }^2\left(\frac{x}{2}\right)}{2{\tan }^2\left(\frac{x}{2}\right)+4\tan \left(\frac{x}{2}\right)+4}dx$

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

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

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

$={\left[\left({\tan }^{-1}(\tan \frac{x}{2}+1)\right)\right]}_0^{\pi }$

$={\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(1\right)$

$=\frac{\pi }{2}-\frac{\pi }{4}$

$=\frac{\pi }{4}$