JEE Mathematics 2024 Question with Solution

Given: the solution states the function in integral form as

$$y(x)=\int \frac{\csc x + \sin x}{\csc x \, \sec x + \tan x \, \sin^2 x} \, dx$$

with the condition

$$\lim_{x \to -\frac{\pi}{2}} y(x)=0$$

Find: $y\left(\frac{\pi}{4}\right)$.

Simplify the integrand using

$$\csc x = \frac{1}{\sin x}, \qquad \sec x = \frac{1}{\cos x}, \qquad \tan x = \frac{\sin x}{\cos x}$$

Then

$$\frac{\csc x + \sin x}{\csc x \, \sec x + \tan x \, \sin^2 x} = \frac{\frac{1}{\sin x}+\sin x}{\frac{1}{\sin x \cos x}+\frac{\sin^3 x}{\cos x}} = \frac{(1+\sin^2 x)\cos x}{1+\sin^4 x}$$

Now use the substitution

$$t=\sin x, \qquad dt = \cos x \, dx$$

So the integral becomes

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

As stated in the solution, this evaluates to

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

Substitute back $t=\sin x$ and apply the given condition from the solution. Then at $x=\frac{\pi}{4}$, we have

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

Hence, as concluded in the provided solution,

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

Therefore, the correct option is D.

Given:

$$y(x)=\int \frac{\csc x + \sin x}{\csc x \, \sec x + \tan x \, \sin^2 x} \, dx$$

with

$$\lim_{x \to -\frac{\pi}{2}} y(x)=0$$

Find: $y\left(\frac{\pi}{4}\right)$.

Rewrite the denominator:

$$\csc x \, \sec x + \tan x \, \sin^2 x = \frac{1}{\sin x \cos x} + \frac{\sin^3 x}{\cos x} = \frac{1+\sin^4 x}{\sin x \cos x}$$

Also,

$$\csc x + \sin x = \frac{1}{\sin x}+\sin x = \frac{1+\sin^2 x}{\sin x}$$

Therefore,

$$\frac{\csc x + \sin x}{\csc x \, \sec x + \tan x \, \sin^2 x} = \frac{\frac{1+\sin^2 x}{\sin x}}{\frac{1+\sin^4 x}{\sin x \cos x}} = \frac{(1+\sin^2 x)\cos x}{1+\sin^4 x}$$

Let

$$t=\sin x$$

Then

$$dt=\cos x \, dx$$

and so

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

the solution states that this gives the required value at $x=\frac{\pi}{4}$ as

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

Note: The given question text appears inconsistent with the solution, which treats $y(x)$ as an integral-defined function. Since the solution is the primary source for answer resolution and explicitly concludes option D, we accept

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

as the final answer.