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

Correct Answer: 8

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: 
\[ a + b = 2 + 6 = 8 \]