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

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(D) Given the integral $y(x) = \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x}} \, dx$. Simplifying the integr

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

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(D) Given the integral $y(x) = \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x}} \, dx$. Simplifying the integrand: $y(x) = \int \frac{\frac{1+\sin^2 x}{\sin x}}{\frac{1+\sin^4 x}{\sin x \cos x}} \, dx = \int \frac{(1+\sin^2 x) \cos x}{1+\sin^4 x} \, dx$. Let $\sin x = t$,then $\cos x \, dx = dt$. $y(t) = \int \frac{1+t^2}{1+t^4} \, dt = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \, dt = \int \frac{1 + \frac{1}{t^2}}{(t - \frac{1}{t})^2 + 2} \, dt$. Let $u = t - \frac{1}{t}$,then $du = (1 + \frac{1}{t^2}) \, dt$. $y(u) = \int \frac{du}{u^2 + 2} = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{u}{\sqrt{2}}ight) + C = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{t - \frac{1}{t}}{\sqrt{2}}ight) + C$. As $x ightarrow \frac{\pi}{2}$,$t ightarrow 1$,so $u ightarrow 0$. Given $\lim_{x ightarrow (\frac{\pi}{2})^-} y(x) = 0$,we have $\frac{1}{\sqrt{2}} 	an^{-1}(0) + C = 0 \implies C = 0$. At $x = \frac{\pi}{4}$,$t = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. $u = \frac{1}{\sqrt{2}} - \sqrt{2} = \frac{1-2}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$. Thus,$y\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}ight) = \frac{1}{\sqrt{2}} 	an^{-1}\left(-\frac{1}{2}ight)$.

For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,if $y(x) = \int \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x} \, dx$ and $\lim_{x \rightarrow (\frac{\pi}{2})^-} y(x) = 0$,then $y\left(\frac{\pi}{4}\right)$ is equal to:

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