x in left(-frac{pi}{2}, frac{pi}{2}right) के | vedclass

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(D) दिया गया समाकलन $y(x) = \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x}} \, dx$ है। समाकल्य को सरल करने पर

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} 	an^{-1}\left(-\frac{1}{2}ight)$

Vedclass · Answer

(D) दिया गया समाकलन $y(x) = \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x}} \, dx$ है। समाकल्य को सरल करने पर: $y(x) = \int \frac{\frac{1+\sin^2 x}{\sin x}}{\frac{1+\sin^4 x}{\sin x \cos x}} \, dx = \int \frac{(1+\sin^2 x) \cos x}{1+\sin^4 x} \, dx$. माना $\sin x = t$,तब $\cos x \, dx = dt$. $y(t) = \int \frac{1+t^2}{1+t^4} \, dt = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \, dt = \int \frac{1 + \frac{1}{t^2}}{(t - \frac{1}{t})^2 + 2} \, dt$. माना $u = t - \frac{1}{t}$,तब $du = (1 + \frac{1}{t^2}) \, dt$. $y(u) = \int \frac{du}{u^2 + 2} = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{u}{\sqrt{2}}ight) + C = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{t - \frac{1}{t}}{\sqrt{2}}ight) + C$. जैसे $x ightarrow \frac{\pi}{2}$,$t ightarrow 1$,इसलिए $u ightarrow 0$. दिया है $\lim_{x ightarrow (\frac{\pi}{2})^-} y(x) = 0$,अतः $\frac{1}{\sqrt{2}} 	an^{-1}(0) + C = 0 \implies C = 0$. $x = \frac{\pi}{4}$ पर,$t = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. $u = \frac{1}{\sqrt{2}} - \sqrt{2} = \frac{1-2}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$. अतः,$y\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}ight) = \frac{1}{\sqrt{2}} 	an^{-1}\left(-\frac{1}{2}ight)$.

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