int_{-frac{pi}{2}}^{frac{pi}{2}} frac{x^2 cos | vedclass

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Question

(A) ધારો કે $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} dx$ ... $(i)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$

Vedclass · Accepted Answer

$\frac{\pi^2}{4}-2$

Vedclass · Answer

(A) ધારો કે $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} dx$ ... $(i)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા,અહીં $a+b = 0$ છે.$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x \cdot e^x}{e^x+1} dx$ ... (ii)$(i)$ અને (ii) નો સરવાળો કરતા:$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x (1+e^x)}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x dx$$x^2 \cos x$ એ યુગ્મ વિધેય હોવાથી,$2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$ થાય.તેથી $I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$.ખંડશઃ સંકલનનો ઉપયોગ કરતા:$I = [x^2 \sin x]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x dx$$I = (\frac{\pi^2}{4} - 0) - 2 [x(-\cos x) - \int 1(-\cos x) dx]_{0}^{\frac{\pi}{2}}$$I = \frac{\pi^2}{4} - 2 [-x \cos x + \sin x]_{0}^{\frac{\pi}{2}}$$I = \frac{\pi^2}{4} - 2 [1 - 0] = \frac{\pi^2}{4} - 2$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} d x$ નું મૂલ્ય કેટલું થાય?

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