int_0^{frac{pi}{2}} frac{cos x , dx}{sqrt{1+c | vedclass

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(C) ધારો કે $I = \int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} \quad \dots (i)$ ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \,

Vedclass · Accepted Answer

$\sqrt{2} \sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Vedclass · Answer

(C) ધારો કે $I = \int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} \quad \dots (i)$ ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા,આપણને મળે: $I = \int_0^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-x) \, dx}{\sqrt{1+\cos(\frac{\pi}{2}-x) \sin(\frac{\pi}{2}-x)}} = \int_0^{\frac{\pi}{2}} \frac{\sin x \, dx}{\sqrt{1+\sin x \cos x}} \quad \dots (ii)$ $(i)$ અને $(ii)$ નો સરવાળો કરતા: $2I = \int_0^{\frac{\pi}{2}} \frac{\cos x + \sin x}{\sqrt{1+\cos x \sin x}} \, dx$ અંશ અને છેદને $\sqrt{2}$ વડે ગુણતા: $2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x + \sin x)}{\sqrt{2+2\sin x \cos x}} \, dx = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x + \sin x)}{\sqrt{3 - (1 - 2\sin x \cos x)}} \, dx$ કારણ કે $1 - 2\sin x \cos x = (\sin x - \cos x)^2$ હોવાથી: $2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x + \sin x)}{\sqrt{3 - (\sin x - \cos x)^2}} \, dx$ ધારો કે $t = \sin x - \cos x$,તો $dt = (\cos x + \sin x) \, dx$. સીમાઓ: જ્યારે $x=0, t=-1$; જ્યારે $x=\frac{\pi}{2}, t=1$. $2I = \int_{-1}^1 \frac{\sqrt{2} \, dt}{\sqrt{3-t^2}} = 2 \int_0^1 \frac{\sqrt{2} \, dt}{\sqrt{(\sqrt{3})^2 - t^2}}$ $I = \sqrt{2} \left[ \sin^{-1} \left( \frac{t}{\sqrt{3}} \right) \right]_0^1 = \sqrt{2} \sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$

$\int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} = $

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