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Answer: $\frac{\pi}{4}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: $\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx$

Solution:$I=\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx$

Put $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}, \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

$I=\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx$

$=\int_{0}^{\pi} \frac{1}{3+2 \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$

$=\int_{0}^{\pi} \frac{1}{\frac{3\left(1+\tan ^{2} \frac{x}{2}\right)+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$

\begin{aligned} &=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{3+3 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \\ &=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{4+2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sec ^{2} x=1+\tan ^{2} x\right] \end{aligned}

Put $\tan \frac{x}{2}=t$

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

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

When x=0  then t=0  and when $x =\pi$   then $t=\infty$

\begin{aligned} &I=\int_{0}^{\infty} \frac{1}{4+2 t^{2}+4 t} 2 d t \\ &=2 \times \frac{1}{2} \int_{0}^{\infty} \frac{1}{2+t^{2}+2 t} d t \\ &=\int_{0}^{\infty} \frac{1}{t^{2}+2 t+1+1} d t \\ &=\int_{0}^{\infty} \frac{1}{\left(t^{2}+2 t+1\right)+1} d t \\ &=\int_{0}^{\infty} \frac{1}{(t+1)^{2}+1^{2}} d t \end{aligned}

\begin{aligned} &=\frac{1}{1}\left[\tan ^{-1}(t+1)\right]_{0}^{\infty} \\ &=\tan ^{-1}(\infty+1)-\tan ^{-1}(0+1) \\ &=\tan ^{-1} \infty-\tan ^{-1} 1 \\ &=\frac{\pi}{2}-\frac{\pi}{4} \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} \infty=\frac{\pi}{2}, \tan ^{-1} 1=\frac{\pi}{4}\right] \\ &=\frac{2 \pi-\pi}{4} \\ &=\frac{\pi}{4} \end{aligned}

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