Integration
Indefinite Integration
Grade Class 12
Question:
For $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, if $y(x) = \int \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x} dx$ and $\lim_{x \to \frac{\pi}{2}} y(x) = 0$ then $y\left( \frac{\pi}{4} \right)$ is equal to
(1) $\tan^{-1}\left( \frac{1}{\sqrt{2}} \right)$
(2) $\frac{1}{2} \tan^{-1}\left( \frac{1}{\sqrt{2}} \right)$
(3) $-\frac{1}{2} \tan^{-1}\left( \frac{1}{\sqrt{2}} \right)$
(4) $\frac{1}{\sqrt{2}} \tan^{-1}\left( -\frac{1}{2} \right)$
Step-by-Step Solution
Key Concept: Simplify the integrand by expressing it in terms of sin x and cos x, then use substitution to evaluate the integral.
The integrand is $\frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x \cos x} + \frac{\sin x}{\cos x} \sin^2 x} = \frac{\frac{1+\sin^2 x}{\sin x}}{\frac{1+\sin^4 x}{\sin x \cos x}} = \frac{\cos x (1+\sin^2 x)}{1+\sin^4 x}$. Let $u = \sin x$, then $du = \cos x dx$. The integral becomes $\int \frac{1+u^2}{1+u^4} du = \int \frac{1+1/u^2}{u^2+1/u^2} du = \int \frac{1+1/u^2}{(u-1/u)^2+2} du$. Let $t = u-1/u$, then $dt = (1+1/u^2) du$. The integral is $\int \frac{dt}{t^2+2} = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{t}{\sqrt{2}} \right) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{\sin x - \operatorname{cosec} x}{\sqrt{2}} \right) + C$. Using the limit condition, we find $C$. Evaluating at $\pi/4$ gives the result.
Correct Answer: (4)