Question

If $\int_{0}^{\pi/4} \frac{\sin^2 x}{1 + \sin x \cos x} \, dx = \frac{1}{a} \log_e \left( \frac{a}{3} \right) + \frac{\pi}{b\sqrt{3}},$where $a, b \in \mathbb{N}$, then $a + b$ is equal to _____

Answer 1

Given integral:
$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2 x}{1 + \sin x \cos x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sin^2 x}{1 + \frac{1}{2} \sin 2x} dx = \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos 2x}{2 + \sin 2x} dx$

We separate this into two integrals:
$\int_{0}^{\frac{\pi}{4}} \frac{1}{2 + \sin 2x} dx - \int_{0}^{\frac{\pi}{4}} \frac{\cos 2x}{2 + \sin 2x} dx$

Denote these integrals as $I_1$ and $I_2$ respectively:
$I_1 = \int_{0}^{\frac{\pi}{4}} \frac{dx}{2 + \sin 2x}, \quad I_2 = \int_{0}^{\frac{\pi}{4}} \frac{\cos 2x}{2 + \sin 2x} dx$

For $I_1$, let $\tan x = t$, hence:
$dx = \frac{dt}{1 + t^2}, \quad \sin 2x = \frac{2t}{1 + t^2}$

Substituting these into the integral:
$I_1 = \frac{1}{2} \int_{0}^{1} \frac{dt}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{\pi}{6\sqrt{3}}$
For $I_2$, we use:
$I_2 = \int_{0}^{\frac{\pi}{4}} \frac{\cos 2x}{2 + \sin 2x} dx$
Applying another substitution and evaluating, we find:
$I_2 = \frac{1}{2} \ln \left(\frac{3}{2}\right)$

Thus, the original integral becomes:
$I = I_1 - I_2 = \frac{\pi}{6\sqrt{3}} - \frac{1}{2} \ln \left(\frac{3}{2}\right)$

Given that:
$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2 x}{1 + \sin x \cos x} dx = \frac{1}{a} \ln_e \left(\frac{a}{3}\right) + \frac{\pi}{b\sqrt{3}}$
Comparing terms, we find: $a = 2, \quad b = 6$
Therefore:
$a + b = 2 + 6 = 8$