The value of the integral int_0^{frac{pi}{4}} | vedclass

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(C) Let $I = \int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}$.Substitute $2x = t$,so $dx = \frac{1}{2} dt$. When $x=0, t=0$ and when $x

Vedclass · Accepted Answer

$\frac{\sqrt{2} \pi^2}{32}$

Vedclass · Answer

(C) Let $I = \int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}$.Substitute $2x = t$,so $dx = \frac{1}{2} dt$. When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=\frac{\pi}{2}$.$I = \int_0^{\frac{\pi}{2}} \frac{(t/2) \cdot (1/2) dt}{\sin^4 t + \cos^4 t} = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{t \, dt}{\sin^4 t + \cos^4 t}$.Using the property $\int_0^a f(t) dt = \int_0^a f(a-t) dt$,we get:$I = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - t) dt}{\cos^4 t + \sin^4 t} = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t} - I$.Thus,$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}$.Divide numerator and denominator by $\cos^4 t$:$2I = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{\sec^4 t \, dt}{	an^4 t + 1} = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} \frac{(1 + 	an^2 t) \sec^2 t \, dt}{	an^4 t + 1}$.Let $	an t = y$,then $\sec^2 t \, dt = dy$:$2I = \frac{\pi}{8} \int_0^{\infty} \frac{1 + y^2}{y^4 + 1} dy = \frac{\pi}{8} \int_0^{\infty} \frac{1 + 1/y^2}{y^2 + 1/y^2} dy$.Let $y - 1/y = p$,then $(1 + 1/y^2) dy = dp$. Limits change from $-\infty$ to $\infty$:$2I = \frac{\pi}{8} \int_{-\infty}^{\infty} \frac{dp}{p^2 + 2} = \frac{\pi}{8} \left[ \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{p}{\sqrt{2}} ight) ight]_{-\infty}^{\infty} = \frac{\pi}{8\sqrt{2}} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{\pi^2}{8\sqrt{2}}$.Therefore,$I = \frac{\pi^2}{16\sqrt{2}} = \frac{\sqrt{2}\pi^2}{32}$.

The value of the integral $\int_0^{\frac{\pi}{4}} \frac{x \, dx}{\sin^4(2x) + \cos^4(2x)}$ equals :

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