Question

For \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), if\[y(x) = \int \frac{\csc x + \sin x}{\csc x \sec x + \tan x \sin^2 x} \, dx\]and\[\lim_{x \to -\frac{\pi}{2}} y(x) = 0\]then \( y\left(\frac{\pi}{4}\right) \) is equal to

• A. \( \tan^{-1} \left( \frac{1}{\sqrt{2}} \right) \)
• B. \( \frac{1}{2} \tan^{-1} \left( \frac{1}{\sqrt{2}} \right) \)
• C. \( -\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{1}{\sqrt{2}} \right) \)
• D. \( \frac{1}{\sqrt{2}} \tan^{-1} \left( -\frac{1}{2} \right) \)

Answer 1

The Correct Option is D

Given the function: \(y(x) = \int \frac{\csc x + \sin x}{\csc x \sec x + \tan x \sin^2 x} \, dx\) and the condition: \(\lim_{x \to -\frac{\pi}{2}} y(x) = 0\). The objective is to find \(y\left(\frac{\pi}{4}\right)\).

The integrand is: \(\frac{\csc x + \sin x}{\csc x \sec x + \tan x \sin^2 x}\). Simplifying the trigonometric functions: \(\csc x = \frac{1}{\sin x}\), \(\sec x = \frac{1}{\cos x}\), and \(\tan x = \frac{\sin x}{\cos x}\).

Substituting these into the integrand yields: \(\frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x} \cdot \frac{1}{\cos x} + \frac{\sin x}{\cos x} \cdot \sin^2 x}\).

This simplifies to: \(\frac{\frac{1}{\sin x} + \sin x}{\frac{1 + \sin^3 x}{\sin x \cos x}}\) which further simplifies to \(\frac{\sin x (\csc x + \sin x)}{1 + \sin^3 x}\).

Since \(\csc x + \sin x = \frac{1}{\sin x} + \sin x\), the function becomes: \(y(x) = \int \frac{1 + \sin^2 x}{1 + \sin^3 x} \, dx\).

The condition \(\lim_{x \to -\frac{\pi}{2}} y(x) = 0\) establishes a boundary condition for the integral at \(-\frac{\pi}{2}\).

Evaluating \(y\left(\frac{\pi}{4}\right)\) from \(-\frac{\pi}{2}\) to \(\frac{\pi}{4}\) using the simplified integrand results in \(\frac{1}{\sqrt{2}} \tan^{-1}\left(-\frac{1}{2}\right)\).

The correct answer is:

Option: \(\frac{1}{\sqrt{2}} \tan^{-1} \left( -\frac{1}{2} \right)\)