The value of the integral int_{frac{pi}{4}}^{ | vedclass

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(B) Let $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1+\sin x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where

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$\pi(\sqrt{2}-1)$

Vedclass · Answer

(B) Let $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1+\sin x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$: $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi-x}{1+\sin(\pi-x)} dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi-x}{1+\sin x} dx$. Adding the two expressions for $I$: $2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x + \pi - x}{1+\sin x} dx = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1+\sin x} dx$. Multiply numerator and denominator by $(1-\sin x)$: $2I = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1-\sin x}{\cos^2 x} dx = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (\sec^2 x - \sec x 	an x) dx$. Integrating: $2I = \pi [	an x - \sec x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$. Evaluating at the limits: $2I = \pi [(	an \frac{3\pi}{4} - \sec \frac{3\pi}{4}) - (	an \frac{\pi}{4} - \sec \frac{\pi}{4})]$. $2I = \pi [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] = \pi [\sqrt{2} - 1 - 1 + \sqrt{2}] = \pi [2\sqrt{2} - 2]$. $I = \pi(\sqrt{2}-1)$.

The value of the integral $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1+\sin x} dx$ is

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