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$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ is equal to
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Verified Answer
The correct answer is:
$2 \sin \frac{\pi}{2}$
From Eqs. (i) and (ii),
$$
\begin{aligned}
\Rightarrow 2 I & =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1-\cos x+1+\cos x}{(1+\cos x)(1-\cos x)} d x \\
\Rightarrow 2 I & =2 \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1}{1-\cos ^2 x} d x=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \operatorname{cosec}^2 x d x \\
\Rightarrow I & =-[\cot x]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}=-\left[\cot \left(\frac{3 \pi}{4}\right)-\cot \left(\frac{\pi}{4}\right)\right] \\
\Rightarrow I & =-\left[-\cot \frac{\pi}{4}-\cot \frac{\pi}{4}\right]=2 \cot \frac{\pi}{4}=2.1 \\
& I=2=2 \sin \frac{\pi}{2}
\end{aligned}
$$
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