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

Simplify the Integrand: - Rewrite the integrand as:

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

- Let \( \sin x = t \), so \( \cos x \, dx = dt \).

Substitute and Integrate: - Substituting \( \sin x = t \), we get:

\(y(x) = \int \frac{1 + t^2}{t^4 + 1} \, dt = \frac{1}{\sqrt{2}} \tan^{-1} \left( t - \frac{1}{\sqrt{2}} \right) + C\)

Determine the Constant \( C \): - At \( x = \frac{\pi}{4} \), \( t = \frac{1}{\sqrt{2}} \). - Since \( \lim_{x \to -\frac{\pi}{2}} y(x) = 0 \), we find \( C = 0 \).

Calculate \( y \left( \frac{\pi}{4} \right) \): - For \( x = \frac{\pi}{4} \), \( t = \frac{1}{\sqrt{2}} \). - Thus:

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

So, the correct answer is: \( \frac{1}{\sqrt{2}} \tan^{-1} \left( -\frac{1}{2} \right) \)