int_0^{pi /2} frac{x sin x cos x}{cos^4 x + s | vedclass

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

Vedclass · Accepted Answer

$\frac{\pi^2}{16}$

Vedclass · Answer

(D) ધારો કે $I = \int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$ .....$(i)$ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા:$I = \int_0^{\pi /2} \frac{(\frac{\pi}{2} - x) \cos x \sin x}{\sin^4 x + \cos^4 x} \, dx$ .....$(ii)$$(i)$ અને $(ii)$ નો સરવાળો કરતા:$2I = \int_0^{\pi /2} \frac{\frac{\pi}{2} \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$અંશ અને છેદને $\cos^4 x$ વડે ભાગતા:$I = \frac{\pi}{4} \int_0^{\pi /2} \frac{	an x \sec^2 x}{1 + 	an^4 x} \, dx$ધારો કે $	an^2 x = t$,તેથી $2 	an x \sec^2 x \, dx = dt$,એટલે કે $	an x \sec^2 x \, dx = \frac{1}{2} dt$. જ્યારે $x=0, t=0$; જ્યારે $x=\frac{\pi}{2}, t 	o \infty$.$I = \frac{\pi}{4} \int_0^{\infty} \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt = \frac{\pi}{8} [	an^{-1} t]_0^{\infty} = \frac{\pi}{8} \cdot \frac{\pi}{2} = \frac{\pi^2}{16}$.

$\int_0^{\pi /2} \frac{x \sin x \cos x}{\cos^4 x + \sin^4 x} \, dx = $

Similar Questions

$\int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} = $

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નિશ્ચિત સંકલન $\int_{0}^{\pi/2} \sqrt{\tan x} \, dx$ નું મૂલ્ય શોધો.

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