Sankalp JEE Full Test-4 Question-10 Solution

Question 10. For x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), let y(x)=\int \frac{\cosec x+\sin x}{\cosec x \sec x+\tan x \sin^{2} x}\,dx and \lim_{x \to \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)

Answer (4)

Explanation:

First simplify the integrand,

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

Put \sin x=t

Then dt=\cos x\,dx

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

Integrating,

y(x)=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+C

Use the given condition x \to \frac{\pi}{2}

Then t=\sin x \to 1

So, 0=\frac{1}{\sqrt{2}}\tan^{-1}(0)+C

Hence, C = 0

Now, for x=\frac{\pi}{4}

t=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}

Therefore,

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