int_0^{pi /2} frac{sin^{3/2} x}{cos^{3/2} x + | vedclass

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Question

(D) ધારો કે $I = \int_0^{\pi /2} \frac{\sin^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx$ ..... $(i)$ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપ

Vedclass · Accepted Answer

$\pi /4$

Vedclass · Answer

(D) ધારો કે $I = \int_0^{\pi /2} \frac{\sin^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx$ ..... $(i)$ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા:$I = \int_0^{\pi /2} \frac{\sin^{3/2} (\pi /2 - x)}{\cos^{3/2} (\pi /2 - x) + \sin^{3/2} (\pi /2 - x)} dx$કારણ કે $\sin(\pi /2 - x) = \cos x$ અને $\cos(\pi /2 - x) = \sin x$,તેથી:$I = \int_0^{\pi /2} \frac{\cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx$ ..... $(ii)$$(i)$ અને $(ii)$ નો સરવાળો કરતા:$2I = \int_0^{\pi /2} \frac{\sin^{3/2} x + \cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx$$2I = \int_0^{\pi /2} 1 dx$$2I = [x]_0^{\pi /2} = \pi /2$$I = \pi /4$

$\int_0^{\pi /2} \frac{\sin^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx = $

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