int_{-pi}^pi frac{x sin x}{1+cos^2 x} dx = | vedclass

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Question

(D) ધારો કે $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. અહીં $f(x) = \frac{x \sin x}{1+\cos^2 x}$ એ યુગ્મ વિધેય છે કારણ કે $f(-x) = f(x)$,તે

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) ધારો કે $I = \int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx$. અહીં $f(x) = \frac{x \sin x}{1+\cos^2 x}$ એ યુગ્મ વિધેય છે કારણ કે $f(-x) = f(x)$,તેથી: $I = 2 \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા: $I = 2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$. $2I = 2\pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx \Rightarrow I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. ધારો કે $\cos x = t$,તો $-\sin x dx = dt$. $I = \pi \int_{-1}^1 \frac{dt}{1+t^2} = \pi [	an^{-1} t]_{-1}^1 = \pi (\frac{\pi}{4} + \frac{\pi}{4}) = \frac{\pi^2}{2}$.

$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos^2 x} dx =$

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