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$\int_0^{\pi / 2} \frac{d x}{2+\cos x}=$
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Verified Answer
The correct answer is:
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\begin{aligned} & I=\int_0^{\pi / 2} \frac{d x}{2+\cos x} \\ &=\int_0^{\pi / 2} \frac{d x}{2 \sin ^2 \frac{x}{2}+2 \cos ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}} \\ &=\int_0^{\pi / 2} \frac{d x}{\sin ^2 \frac{x}{2}+3 \cos ^2 \frac{x}{2}}=\int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2}}{3+\tan ^2 \frac{x}{2}} d x \\ & \text { Put } t=\tan \frac{x}{2} \Rightarrow d t=\frac{1}{2} \sec ^2 \frac{x}{2} d x \text {, then }\end{aligned}$
$I=2 \int_0^1 \frac{d t}{3+t^2}=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$I=2 \int_0^1 \frac{d t}{3+t^2}=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
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