Question

\( \int_{0}^{\pi/4} \frac{\cos^2 x \sin^2 x}{\left( \cos^3 x + \sin^3 x \right)^2} \, dx \) is equal to:

• A. \( \frac{1}{12} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{1}{3} \)

Answer 1

The Correct Option is C

Divide the numerator and denominator by \(\cos x\):

\[ \int_0^{\pi/4} \frac{\tan^2 x \sec^2 x \, dx}{(1 + \tan^3 x)^2}. \]

Let \(1 + \tan^3 x = t\). Then:

\[ \tan^2 x \sec^2 x \, dx = \frac{dt}{3}. \]

The limits transform as:

• When \(x = 0\), \(t = 1\),
• and when \(x = \frac{\pi}{4}\), \(t = 2\).

Substitute into the integral:

\[ \int_0^{\pi/4} \frac{\tan^2 x \sec^2 x \, dx}{(1 + \tan^3 x)^2} = \frac{1}{3} \int_1^2 \frac{dt}{t^2}. \]

Solve the integral:

\[ \frac{1}{3} \int_1^2 t^{-2} \, dt = \frac{1}{3} \left[ -\frac{1}{t} \right]_1^2. \]

Simplify:

\[ \frac{1}{3} \left[ -\frac{1}{2} - (-1) \right] = \frac{1}{3} \left[ -\frac{1}{2} + 1 \right] = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}. \]