For x ∈ (- (π)/(2), (π)/(2)) , if y(x) = ∫ cosec x + x cosec xsecx...

Given:

$f\left(x\right) = ∫ \dfrac{cosecx + \sin x}{cosecxsecx + \tan x\sin^{2}x}dx$

$⇒ f\left(x\right) = ∫ \dfrac{\dfrac{1}{\sin x} + \sin x}{\dfrac{1}{\sin x \cos x} + \dfrac{\sin x}{\cos x}\left(\sin^{2}x\right)}dx$

$⇒ f\left(x\right) = ∫ \dfrac{\dfrac{1 + \sin^{2}x}{\sin x}}{\dfrac{1 + \sin^{4}x}{\sin x \cos x}}dx$

$⇒ f\left(x\right) = ∫ \dfrac{\left(1 + \sin^{2}x\right)\cos x}{1 + \sin^{4}x}dx$

Let, $\sin x = t$

$cosxdx = dt$

$⇒ f\left(x\right) = ∫ \dfrac{1 + {t}^{2}}{1 + {t}^{4}}dt$

$⇒ f\left(x\right) = ∫ \dfrac{1 + \dfrac{1}{{t}^{2}}}{{t}^{2} + \dfrac{1}{{t}^{2}}}dt$

$⇒ f\left(x\right) = ∫ \dfrac{1 + \dfrac{1}{{t}^{2}}}{{\left(t - \dfrac{1}{t}\right)}^{2} + 2}dt$

Let, $t - \dfrac{1}{t} = u$

$⇒ f\left(x\right) = ∫ \dfrac{1}{{u}^{2} + 2}du$

$⇒ f\left(x\right) = \dfrac{1}{\sqrt{2}}\tan^{- 1}\left(\dfrac{u}{\sqrt{2}}\right) + C$

$⇒ f\left(x\right) = \dfrac{1}{\sqrt{2}}\tan^{- 1}\left(\dfrac{t - \dfrac{1}{t}}{\sqrt{2}}\right) + C$

$⇒ f\left(x\right) = \dfrac{1}{\sqrt{2}}\tan^{- 1}\left(\dfrac{\sin x - \dfrac{1}{\sin x}}{\sqrt{2}}\right) + C$

$⇒ f\left(x\right) = \dfrac{1}{\sqrt{2}}\tan^{- 1}\left(\dfrac{\sin x - cosecx}{\sqrt{2}}\right) + C$

Now, $\lim_{x \to \dfrac{\pi}{2} -}f\left(x\right) = 0$

$⇒ \lim_{x \to \dfrac{\pi}{2} -}f\left(x\right) = \lim_{x \to \dfrac{\pi}{2} -}\left[\dfrac{1}{\sqrt{2}}\tan^{- 1}\left(\dfrac{\sin x - cosecx}{\sqrt{2}}\right) + C\right] = 0$

$⇒ \dfrac{1}{\sqrt{2}}\tan^{- 1}\left(0\right) + C = 0$

$⇒ C = 0$

$⇒ f\left(x\right) = \dfrac{1}{\sqrt{2}}\tan^{- 1}\left(\dfrac{\sin x - cosecx}{\sqrt{2}}\right)$

$⇒ y\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}\tan^{- 1}\left(\dfrac{sin\dfrac{\pi}{4} - cosec\dfrac{\pi}{4}}{\sqrt{2}}\right)$

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

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

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