A tricky trigonometric integral

Split the integral into 3 integrals, and denote them as $A,B,C$ , respectively. The latter two integrals can be expressed as $\displaystyle \int c \cdot \dfrac{f'(x)}{f(x)} \, dx$ for some constant $c$ . Using the property $\frac d{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$ . We can now evaluate $B$ to be:

$B = 9 \sqrt2 \int_0^{\pi/3} \dfrac{ \cos(3x/4)}{2\sin(3x/4) + 1} \, dx \stackrel{y = 3x/4}{=} 9 \sqrt2 \cdot \frac43 \cdot \ln(1 + \sqrt2) = 6\sqrt2 \ln(1 + \sqrt2)$

Analogously, $C = 4 \ln(\frac53)$ .

We are left to solve the most tedious integral, $A$ . In the intermediate step below, we will be using the double angle formula, $\tan (2X) = \dfrac{\sin(2X)}{\cos(2X)} = \dfrac{2\sin X \cos X}{1 - 2\sin^2 X}$ . $A = \int_0^{\pi /3} \dfrac{\tan^2 (x)}{\cos^3 (x/2)} \, dx \stackrel{y =x/2}{=} 2 \int_0^{\pi/6} \left( \dfrac{2\sin y \cos y}{1 - 2\sin^2 y}\right)^2 \cdot \frac1{\cos^3 y} \, dy = 8 \int_0^{\pi/6} \dfrac{\sin^2 y}{\cos y} \cdot \frac1{(1-2\sin^2 y)^2} \, dy \stackrel{z = \sin y}{=} \int_0^{1/2} \dfrac{z^2}{1-z^2} \cdot \frac8{(1-2z^2)^2} \, dz.$ By partial fraction decomposition, the integrand in the last integral above is equal to $\dfrac{16}{2z^2 - 1} + \dfrac8{(2z^2 - 1)^2} + \dfrac4{z+1} - \dfrac4{z-1}.$ Going through all the tedious algebra, we get $A = 4 + 4 \ln3 - 3\sqrt2 \ln(3 + 2\sqrt2) .$

Hence, the integral in question is equal to $A+B+C=4 (\ln5 + 1)$ . The answer is $\boxed4$ .