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

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Question

(D) Let $I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{1+\cos x} \quad \dots (i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-

Vedclass · Accepted Answer

$2 \sin \frac{\pi}{2}$

Vedclass · Answer

(D) Let $I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{1+\cos x} \quad \dots (i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we have $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$.So, $I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{1+\cos(\pi-x)} = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{dx}{1-\cos x} \quad \dots (ii)$Adding $(i)$ and $(ii)$:$2I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \left( \frac{1}{1+\cos x} + \frac{1}{1-\cos x} \right) dx$$2I = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1-\cos x + 1+\cos x}{1-\cos^2 x} dx = \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{2}{\sin^2 x} dx$$2I = 2 \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \operatorname{cosec}^2 x dx$$I = [-\cot x]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}$$I = -(\cot \frac{3\pi}{4} - \cot \frac{\pi}{4}) = -(-1 - 1) = 2$Since $2 \sin \frac{\pi}{2} = 2(1) = 2$, the correct option is $D$.

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

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