Indefinite Integration Question 2
Question 2 - 2024 (29 Jan Shift 1)
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 $\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$ then $y\left(\frac{\pi}{4}\right)$ is equal to
(1) $\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
(2) $\frac{1}{2} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
(3) $-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
(4) $\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)$
Show Answer
Answer (3)
Solution
$y(x)=\int \frac{\left(1+\sin ^{2} x\right) \cos x}{1+\sin ^{4} x} d x$
Put $\sin x=t$
$=\int \frac{1+t^{2}}{t^{4}+1} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C$
$x=\frac{\pi}{2}, t=1 \quad \therefore C=0$
$y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
