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

Answer: (C)

$\frac{1}{6}$

Answer 2

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:

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