int_{-frac{pi}{4}}^{frac{pi}{4}} left( frac{x | vedclass

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(D) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2x} dx$. We can split this into two integrals: $I = \int_{-\frac{\pi}

Vedclass · Accepted Answer

$\frac{\pi^2}{6\sqrt{3}}$

Vedclass · Answer

(D) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2x} dx$. We can split this into two integrals: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2-\cos 2x} dx + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{2-\cos 2x} dx$. Let $f(x) = \frac{x}{2-\cos 2x}$. Since $f(-x) = \frac{-x}{2-\cos(-2x)} = -f(x)$,$f(x)$ is an odd function. Thus,$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) dx = 0$. Now,$I = \frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx$. Since the integrand is an even function,$I = 2 	imes \frac{\pi}{4} \int_{0}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx$. Using $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$,we get $I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1+	an^2 x}{2(1+	an^2 x) - (1-	an^2 x)} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 x}{1+3	an^2 x} dx$. Let $t = 	an x$,then $dt = \sec^2 x dx$. Limits change from $[0, \frac{\pi}{4}]$ to $[0, 1]$. $I = \frac{\pi}{2} \int_{0}^{1} \frac{dt}{1+3t^2} = \frac{\pi}{2} \left[ \frac{1}{\sqrt{3}} 	an^{-1}(\sqrt{3}t) ight]_{0}^{1} = \frac{\pi}{2\sqrt{3}} 	an^{-1}(\sqrt{3}) = \frac{\pi}{2\sqrt{3}} 	imes \frac{\pi}{3} = \frac{\pi^2}{6\sqrt{3}}$.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{x+\frac{\pi}{4}}{2-\cos 2x} \right) dx$ is equal to

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