The value of the integral int limits_{-frac{p | vedclass

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(D) Let $I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$  $(1)$Using the property $\int_a^b f(x) dx = \int_a

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

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(D) Let $I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$  $(1)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a+b = 0$,we replace $x$ with $-x$:$I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2(-x)} d x = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2 x} d x$  $(2)$Adding $(1)$ and $(2)$:$2I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(x+\frac{\pi}{4}) + (-x+\frac{\pi}{4})}{2-\cos 2 x} d x = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{2}}{2-\cos 2 x} d x$Since the integrand is an even function,$2I = 2 \cdot \frac{\pi}{2} \int \limits_0^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x = \pi \int \limits_0^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x$$I = \frac{\pi}{2} \int \limits_0^{\frac{\pi}{4}} \frac{\sec^2 x}{2(1+	an^2 x) - (1-	an^2 x)} dx = \frac{\pi}{2} \int \limits_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1+3	an^2 x} dx$Let $t = 	an x$,then $dt = \sec^2 x dx$. When $x=0, t=0$; when $x=\frac{\pi}{4}, t=1$:$I = \frac{\pi}{2} \int \limits_0^1 \frac{dt}{1+3t^2} = \frac{\pi}{2 \cdot 3} \int \limits_0^1 \frac{dt}{\frac{1}{3}+t^2} = \frac{\pi}{6} \cdot \sqrt{3} [	an^{-1}(\sqrt{3}t)]_0^1 = \frac{\pi \sqrt{3}}{6} 	an^{-1}(\sqrt{3}) = \frac{\pi \sqrt{3}}{6} \cdot \frac{\pi}{3} = \frac{\pi^2}{6 \sqrt{3}}$

The value of the integral $\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ is :

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