int_{frac{pi}{4}}^{frac{3 pi}{4}} frac{d x}{1 | vedclass

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Question

(C) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Therefore,the integral becomes $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{2 \cos^2 \frac{x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Therefore,the integral becomes $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{2 \cos^2 \frac{x}{2}} = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \sec^2 \frac{x}{2} dx$.Integrating $\sec^2 \frac{x}{2}$ gives $2 	an \frac{x}{2}$.So,$\frac{1}{2} [2 	an \frac{x}{2}]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} = [	an \frac{x}{2}]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}$.Substituting the limits: $	an \frac{3 \pi}{8} - 	an \frac{\pi}{8}$.Using $	an 	heta - 	an \phi = \frac{\sin(	heta - \phi)}{\cos 	heta \cos \phi}$,or simply $	an \frac{3 \pi}{8} = \cot \frac{\pi}{8}$.Then $\cot \frac{\pi}{8} - 	an \frac{\pi}{8} = \frac{\cos \frac{\pi}{8}}{\sin \frac{\pi}{8}} - \frac{\sin \frac{\pi}{8}}{\cos \frac{\pi}{8}} = \frac{\cos^2 \frac{\pi}{8} - \sin^2 \frac{\pi}{8}}{\sin \frac{\pi}{8} \cos \frac{\pi}{8}}$.Using double angle formulas: $\cos^2 	heta - \sin^2 	heta = \cos 2	heta$ and $2 \sin 	heta \cos 	heta = \sin 2	heta$.This simplifies to $\frac{\cos \frac{\pi}{4}}{\frac{1}{2} \sin \frac{\pi}{4}} = 2 \cot \frac{\pi}{4} = 2(1) = 2$.

$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ is equal to

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