int_{-pi}^pi frac{x sin ^3 x}{4-cos ^2 x} d x | vedclass

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Question

(B) माना $I = \int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$. चूँकि $f(x) = \frac{x \sin ^3 x}{4-\cos ^2 x}$ एक सम फलन है (क्योंकि $f(-x) = f(x)

Vedclass · Accepted Answer

$2 \pi\left(1-\frac{3}{4} \log 3\right)$

Vedclass · Answer

(B) माना $I = \int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$. चूँकि $f(x) = \frac{x \sin ^3 x}{4-\cos ^2 x}$ एक सम फलन है (क्योंकि $f(-x) = f(x)$),इसलिए: $I = 2 \int_0^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$ ....$(i)$ गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर: $I = 2 \int_0^\pi \frac{(\pi-x) \sin ^3 x}{4-\cos ^2 x} d x$ ....(ii) $(i)$ और (ii) को जोड़ने पर: $2I = 2 \pi \int_0^\pi \frac{\sin ^3 x}{4-\cos ^2 x} d x \Rightarrow I = \pi \int_0^\pi \frac{(1-\cos^2 x) \sin x}{4-\cos ^2 x} d x$. माना $\cos x = t$,तब $-\sin x dx = dt$. जब $x=0, t=1$; जब $x=\pi, t=-1$. $I = -\pi \int_1^{-1} \frac{1-t^2}{4-t^2} dt = 2\pi \int_0^1 \frac{1-t^2}{4-t^2} dt$. $I = 2\pi \int_0^1 (1 - \frac{3}{4-t^2}) dt = 2\pi [t - \frac{3}{4} \log |\frac{2+t}{2-t}|]_0^1 = 2\pi [1 - \frac{3}{4} \log 3]$.

$\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x=$

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