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

We need to evaluate $$y(x) = \int\frac{\csc x + \sin x}{\csc x\sec x + \tan x\sin^2 x}\,dx$$ with $$\lim_{x \to (\pi/2)^-} y(x) = 0$$.

Simplify the integrand.

Multiply numerator and denominator by $$\sin x\cos x$$:

Numerator:

$$(\csc x + \sin x) \cdot \sin x\cos x = \cos x + \sin^2 x\cos x = \cos x(1 + \sin^2 x)$$

Denominator:

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

So the integrand simplifies to $$\frac{\cos x(1 + \sin^2 x)}{1 + \sin^4 x}$$.

Substitute $$t = \sin x$$, $$dt = \cos x\,dx$$.

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

Evaluate the integral.

Divide numerator and denominator by $$t^2$$:

$$\frac{1 + 1/t^2}{t^2 + 1/t^2} = \frac{1 + 1/t^2}{(t - 1/t)^2 + 2}$$

Let $$u = t - \frac{1}{t}$$, so $$du = \left(1 + \frac{1}{t^2}\right)dt$$.

$$y = \int\frac{du}{u^2 + 2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \csc x}{\sqrt{2}}\right) + C$$

Apply the boundary condition.

As $$x \to \left(\frac{\pi}{2}\right)^-$$: $$\sin x \to 1$$, so $$\sin x - \csc x \to 0$$.

$$\lim_{x \to (\pi/2)^-} y(x) = \frac{1}{\sqrt{2}}\tan^{-1}(0) + C = C = 0$$

So $$C = 0$$.

Evaluate $$y(\pi/4)$$.

At $$x = \frac{\pi}{4}$$: $$\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$, $$\csc\frac{\pi}{4} = \sqrt{2}$$.

$$\sin x - \csc x = \frac{1}{\sqrt{2}} - \sqrt{2} = \frac{1 - 2}{\sqrt{2}} = \frac{-1}{\sqrt{2}}$$

$$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)$$

The correct answer is Option (4): $$\boxed{\frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)}$$.