JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 13)
For $$x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, if $$y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x$$, and $$\lim _\limits{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$$ then $$y\left(\frac{\pi}{4}\right)$$ is equal to
$$-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$$
$$\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$$
$$\frac{1}{2} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$$
$$\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)$$
Explanation
$$\begin{aligned}
& y(x)=\int \frac{\left(1+\sin ^2 x\right) \cos x}{1+\sin ^4 x} d x \\
& \text { Put } \sin x=t \\
& =\int \frac{1+t^2}{t^4+1} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C \\
& x=\frac{\pi}{2}, t=1 \quad \therefore C=0 \\
& y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)
\end{aligned}$$
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