If I = frac{2}{pi} int_{-pi / 4}^{pi / 4} fra | vedclass

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(B) Let $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x

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$4$

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(B) Let $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:$I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{-\sin x})(2 - \cos 2x)}$.Adding the two expressions for $I$:$2I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{1}{2 - \cos 2x} \left( \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{1 + e^{\sin x}} ight) dx = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{2 - \cos 2x}$.Since the integrand is even,$I = \frac{2}{\pi} \int_0^{\pi / 4} \frac{dx}{2 - \cos 2x} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{2(1 + 	an^2 x) - (1 - 	an^2 x)} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{1 + 3 	an^2 x}$.Let $u = \sqrt{3} 	an x$,then $du = \sqrt{3} \sec^2 x dx$.$I = \frac{2}{\pi \sqrt{3}} \int_0^{\sqrt{3}} \frac{du}{1 + u^2} = \frac{2}{\pi \sqrt{3}} [	an^{-1} u]_0^{\sqrt{3}} = \frac{2}{\pi \sqrt{3}} \cdot \frac{\pi}{3} = \frac{2}{3\sqrt{3}}$.Thus,$27 I^2 = 27 \cdot \frac{4}{27} = 4$.

If $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$,then $27 I^2$ equals . . . . . . . .

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