left|frac{120}{pi^3} int_0^pi frac{x^2 sin x | vedclass

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(A) ધારો કે $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,$I = \int

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) ધારો કે $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,$I = \int_0^\pi \frac{(\pi-x)^2 \sin(\pi-x) \cos(\pi-x)}{\sin^4(\pi-x) + \cos^4(\pi-x)} dx = -\int_0^\pi \frac{(\pi^2 - 2\pi x + x^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. બંને સમીકરણોનો સરવાળો કરતા: $2I = \int_0^\pi \frac{(2\pi x - \pi^2) \sin x \cos x}{\sin^4 x + \cos^4 x} dx$. આને ઉકેલતા,$I = -\frac{\pi^2}{2} \int_0^{\pi/2} \frac{\sin t \cos t}{\sin^4 t + \cos^4 t} dt$. $u = \sin^2 t$ આદેશ લેતા,સંકલન $-\frac{\pi^3}{8}$ મળે છે. માનાંક લેતા,$\left| -\frac{\pi^3}{8} \right| = \frac{\pi^3}{8}$. અંતિમ જવાબ: $\frac{120}{\pi^3} \cdot \frac{\pi^3}{8} = 15$.

$\left|\frac{120}{\pi^3} \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx\right|$ ની કિંમત શોધો.

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