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

Answer: (D)

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

Answer 2

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