int_{frac{pi}{3}}^{frac{2 pi}{3}} frac{x}{1+s | vedclass

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Question

(B) Let $I = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,wher

Vedclass · Accepted Answer

$\pi(2-\sqrt{3})$

Vedclass · Answer

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

$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} \,d x=$

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