Question

If \( \int \frac{\cos^3 x}{\sin^2 x + \sin^4 x} dx = c - \operatorname{cosec} x - f(x) \), then \( f\left(\frac{\pi}{2}\right) = \)

• A. 1
• B. 0
• C. \( \frac{\pi}{2} \)
• D. \( \pi \)

Hint:

1. Simplify the integrand using trigonometric identities (\( \cos^2x = 1-\sin^2x \)). 2. Use substitution (e.g., \(u=\sin x\)). 3. If the resulting rational function in \(u\) is complex, use partial fraction decomposition. 4. Integrate term by term. 5. Compare the result with the given form to identify \(f(x)\). Remember \( \tan^{-1}(1) = \pi/4 \).

Answer 1

The Correct Option is C

Let \( I = \int \frac{\cos^3 x}{\sin^2 x + \sin^4 x} dx = \int \frac{\cos^3 x}{\sin^2 x (1 + \sin^2 x)} dx \).
\[ I = \int \frac{\cos^2 x \cdot \cos x}{\sin^2 x (1 + \sin^2 x)} dx = \int \frac{(1-\sin^2 x)\cos x}{\sin^2 x (1 + \sin^2 x)} dx \] Let \( u = \sin x \).
Then \( du = \cos x dx \).
\[ I = \int \frac{1-u^2}{u^2(1+u^2)} du \] Use partial fractions for \( \frac{1-u^2}{u^2(1+u^2)} \).
Let \( v = u^2 \).
Then \( \frac{1-v}{v(1+v)} = \frac{A}{v} + \frac{B}{1+v} \).
\( 1-v = A(1+v) + Bv \).
If \( v=0 \), \( 1 = A(1) \implies A=1 \).
If \( v=-1 \), \( 1-(-1) = B(-1) \implies 2 = -B \implies B=-2 \).
So, \( \frac{1-u^2}{u^2(1+u^2)} = \frac{1}{u^2} - \frac{2}{1+u^2} \).
\[ I = \int \left( \frac{1}{u^2} - \frac{2}{1+u^2} \right) du = \int u^{-2} du - 2 \int \frac{1}{1+u^2} du \] \[ = \frac{u^{-1}}{-1} - 2\tan^{-1}u + C_0 = -\frac{1}{u} - 2\tan^{-1}u + C_0 \] Substitute back \( u = \sin x \): \[ I = -\frac{1}{\sin x} - 2\tan^{-1}(\sin x) + C_0 \] \[ I = -\operatorname{cosec} x - 2\tan^{-1}(\sin x) + C_0 \] Given \( I = c - \operatorname{cosec} x - f(x) \).
Comparing, \( c = C_0 \) and \( f(x) = 2\tan^{-1}(\sin x) \).
We need to find \( f(\pi/2) \).
\[ f(\pi/2) = 2\tan^{-1}(\sin(\pi/2)) = 2\tan^{-1}(1) \] Since \( \tan^{-1}(1) = \pi/4 \).
\[ f(\pi/2) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \] This matches option (3).