1. Simplify the integrand

Given

$$y(x)=\int \frac{\csc x+\sin x}{\csc x\sec x+\tan x\sin^2x}\,dx$$

for $x\in\left(-\frac\pi2,\frac\pi2\right)$.

Write everything in $\sin x,\cos x$:

$$\csc x=\frac1{\sin x},\qquad \sec x=\frac1{\cos x},\qquad \tan x=\frac{\sin x}{\cos x}.$$

So the numerator becomes

$$\csc x+\sin x=\frac1{\sin x}+\sin x=\frac{1+\sin^2x}{\sin x}.$$

The denominator becomes

$$\csc x\sec x+\tan x\sin^2x =\frac1{\sin x\cos x}+\frac{\sin x}{\cos x}\sin^2x =\frac1{\sin x\cos x}+\frac{\sin^3x}{\cos x}.$$

Taking common denominator:

$$\frac1{\sin x\cos x}+\frac{\sin^3x}{\cos x} =\frac{1+\sin^4x}{\sin x\cos x}.$$

Hence the integrand is

$$\frac{\frac{1+\sin^2x}{\sin x}}{\frac{1+\sin^4x}{\sin x\cos x}} =\frac{(1+\sin^2x)\cos x}{1+\sin^4x}.$$

Therefore,

$$y(x)=\int \frac{(1+\sin^2x)\cos x}{1+\sin^4x}\,dx.$$

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2. Substitute $t=\sin x$

Let

$$t=\sin x \quad\Rightarrow\quad dt=\cos x\,dx.$$

Then

$$y(x)=\int \frac{1+t^2}{1+t^4}\,dt.$$

Now factor:

$$t^4+1=(t^2+\sqrt2 t+1)(t^2-\sqrt2 t+1).$$

A standard decomposition gives

$$\frac{1+t^2}{1+t^4}=\frac12\left(\frac1{t^2+\sqrt2 t+1}+\frac1{t^2-\sqrt2 t+1}\right).$$

So

$$y(x)=\frac12\int\frac{dt}{t^2+\sqrt2 t+1}+\frac12\int\frac{dt}{t^2-\sqrt2 t+1}.$$

Complete squares:

$$t^2+\sqrt2 t+1=\left(t+\frac1{\sqrt2}\right)^2+\frac12,$$ $$t^2-\sqrt2 t+1=\left(t-\frac1{\sqrt2}\right)^2+\frac12.$$

Using

$$\int \frac{du}{u^2+a^2}=\frac1a\tan^{-1}\left(\frac{u}{a}\right),$$

we get

$$\int\frac{dt}{\left(t\pm\frac1{\sqrt2}\right)^2+\frac12} =\sqrt2\tan^{-1}(\sqrt2(t\pm\tfrac1{\sqrt2})) =\sqrt2\tan^{-1}(\sqrt2 t\pm1).$$

Thus

$$y(x)=\frac{\sqrt2}{2}\left[\tan^{-1}(\sqrt2 t+1)+\tan^{-1}(\sqrt2 t-1)\right]+C.$$

Since $t=\sin x$,

$$y(x)=\frac1{\sqrt2}\left[\tan^{-1}(\sqrt2\sin x+1)+\tan^{-1}(\sqrt2\sin x-1)\right]+C.$$

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3. Use the given condition

Given

$$\lim_{x\to(\pi/2)^-}y(x)=0.$$

As $x\to\left(\frac\pi2\right)^-$,

$$\sin x\to 1.$$

So

$$0=\frac1{\sqrt2}\left[\tan^{-1}(\sqrt2+1)+\tan^{-1}(\sqrt2-1)\right]+C.$$

Now use

$$(\sqrt2+1)(\sqrt2-1)=1.$$

For positive $a$,

$$\tan^{-1}a+\tan^{-1}\left(\frac1a\right)=\frac\pi2.$$

Hence

$$\tan^{-1}(\sqrt2+1)+\tan^{-1}(\sqrt2-1)=\frac\pi2.$$

Therefore

$$C=-\frac1{\sqrt2}\cdot\frac\pi2=-\frac{\pi}{2\sqrt2}.$$

So

$$y(x)=\frac1{\sqrt2}\left[\tan^{-1}(\sqrt2\sin x+1)+\tan^{-1}(\sqrt2\sin x-1)-\frac\pi2\right].$$

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4. Compute $y\left(\frac\pi4\right)$

Since

$$\sin\frac\pi4=\frac1{\sqrt2},$$

we get

$$\sqrt2\sin\frac\pi4=1.$$

Thus

$$y\left(\frac\pi4\right)=\frac1{\sqrt2}\left[\tan^{-1}(2)+\tan^{-1}(0)-\frac\pi2\right].$$

Since $\tan^{-1}(0)=0$,

$$y\left(\frac\pi4\right)=\frac1{\sqrt2}\left(\tan^{-1}(2)-\frac\pi2\right).$$

Now

$$\tan^{-1}(2)-\frac\pi2=-\tan^{-1}\left(\frac12\right)$$

because for positive $u$,

$$\tan^{-1}(u)+\tan^{-1}\left(\frac1u\right)=\frac\pi2.$$

Therefore,

$$y\left(\frac\pi4\right)=\frac1{\sqrt2}\tan^{-1}\left(-\frac12\right).$$

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5. Match with options

This is exactly Option D:

$$\boxed{\frac1{\sqrt2}\tan^{-1}\left(-\frac12\right)}.$$