સંકલન int_{-frac{pi}{2}}^{frac{pi}{2}} sin^4 | vedclass

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Question

(C) ધારો કે $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$. ગુણધર્મ $\in

Vedclass · Accepted Answer

$\frac{3}{8}\pi$

Vedclass · Answer

(C) ધારો કે $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$. ગુણધર્મ $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$ નો ઉપયોગ કરતા: $I = \int_{0}^{\frac{\pi}{2}} \left[ \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) + \sin^4(-x) \left( 1 + \log \left( \frac{2 + \sin(-x)}{2 - \sin(-x)} \right) \right) \right] dx$. $\sin(-x) = -\sin x$ હોવાથી: $I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \left[ 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) + 1 + \log \left( \frac{2 - \sin x}{2 + \sin x} \right) \right] dx$. $I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \left[ 2 + \log \left( \frac{2 + \sin x}{2 - \sin x} \cdot \frac{2 - \sin x}{2 + \sin x} \right) \right] dx$. $I = \int_{0}^{\frac{\pi}{2}} \sin^4 x [ 2 + \log(1) ] dx = \int_{0}^{\frac{\pi}{2}} 2 \sin^4 x dx = 2 \int_{0}^{\frac{\pi}{2}} \sin^4 x dx$. વોલિસના સૂત્રનો ઉપયોગ કરતા,$\int_{0}^{\frac{\pi}{2}} \sin^n x dx = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ (જ્યારે $n$ બેકી હોય): $I = 2 \cdot \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \right) = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$.

સંકલન $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$ નું મૂલ્ય શોધો.

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