Party time integral!

So we have,

$\int_0^{\frac{\pi}{4}}\sqrt[3]{\tan(x)} dx$ and let $u^3 = \tan(x)$ or $x = \arctan(x)$ and hence, $dx = \frac{3u^2}{1+u^6} du$ . Also, substituting for x at $0$ and $\frac{\pi}{4}$ gives us the bounds for $u$ .

$\Rightarrow \int_0^1 u\frac{3u^2}{1+(u^2)^3} du$ and let $v = u^2$ to get $dv = 2u du$ or $u du = \frac{1}{2} dv$ . Similarly, substituting at $u = 0, 1$ yields $v = 0, 1$ respectively.

Thus, the integral becomes,

$\Rightarrow \int_0^1 \frac{1}{2} \frac{3v}{v^3+1} dv$

$\Rightarrow \frac{1}{2} \int_0^1 \frac{3v}{(v+1)(v^2-v+1)} dv$

Going into partial fraction decomposition, and taking just the integrand, we have the following prophecy:

$\frac{3v}{(v+1)(v^2-v+1)} = \frac{A}{v+1} + \frac{Bv + C}{v^2-v+1}$

i.e., $3v = A(v^2-v+1) + (Bv + C)(v+1)$

$3v = Av^2 - Av + A + Bv^2 + Bv + Cv + C$

Through pattern matching, we obtain

$v^2: A+B = 0 \Rightarrow B = -A$ , $v: -A+B+C = 3$ , and Constant: $A+C = 0 \Rightarrow C = -A$

Combining that information, $-A-A-A = 3 \Rightarrow A = -1 \Rightarrow B = C = 1$

Therefore,

$\frac{1}{2} \int_0^1 \frac{3v}{(v+1)(v^2-v+1)} dv = \frac{1}{2} \int_0^1 \frac{-1}{v+1} + \frac{v+1}{v^2-v+1} dv$

$\Rightarrow \frac{1}{2}[-ln|v+1|]_0^1 + \frac{1}{2} \int_0^1 \frac{v+1}{v^2-v+1} dv$

And doing a few operations, i.e., multiplying and diving by 2, adding and subtracting 1.

$\Rightarrow \frac{1}{2}(-ln(2)) + \frac{1}{4} \int_0^1 \frac{2v-1+3}{v^-v+1} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{1}{4} \int_0^1 \frac{2v-1}{v^2-v+1} dv + \frac{3}{4} \int_0^1 \frac{1}{v^2-v+1} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{1}{4}[ln(v^2-v+1)]_0^1 + \frac{3}{4} \int_0^1 \frac{1}{v^2-v+1} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{1}{4}[0 - 0] + \frac{3}{4} \int_0^1 \frac{1}{v^2 - v + \frac{1}{4} - \frac{1}{4} + 1} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{3}{4} \int_0^1 \frac{1}{(v-\frac{1}{2})^2 + \frac{3}{4}} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{3}{4} \int_0^1 \frac{1}{\frac{3}{4}(\frac{4}{3}(v-\frac{1}{2})^2 +1} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{3}{4} \int_0^1 \frac{4/3}{(\frac{2}{\sqrt{3}}(v-\frac{1}{2}))^2 + 1} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{3}{4} \int_0^2 \frac{\frac{2}{\sqrt{3}}\frac{2}{\sqrt{3}}}{(\frac{2v}{\sqrt{3}} - \frac{1}{\sqrt{3}})^2 + 1} dv$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{\sqrt{3}}{2} \int_0^1 \frac{2/\sqrt{3}}{(\frac{2v}{\sqrt{3}} - \frac{1}{\sqrt{3}})^2 + 1} dv$

From an awareness/experience in calculus, the $\int \frac{1}{1+x^2} dx = \arctan(x) + C$ , if not, it is quite easy to derive with a trig. substitution of $x = \tan(\theta)$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{\sqrt{3}}{2}[\arctan(\frac{2v}{\sqrt{3}} - \frac{1}{\sqrt{3}})]_0^1$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{\sqrt{3}}{2}(\arctan(\frac{1}{\sqrt{3}} - \arctan(\frac{-1}{\sqrt{3}}))$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{\sqrt{3}}{2}(\frac{\pi}{6} + \frac{\pi}{6})$

$\Rightarrow \frac{-1}{2}ln(2) + \frac{\sqrt{3}}{2}(\frac{\pi}{3})$

We finally arrive at...

$\int_0^{\frac{\pi}{4}} \sqrt[3]{\tan(x)} dx = \frac{\sqrt{3}}{6} \pi - \frac{1}{2}ln(2)$

Therefore, $a = 3$ , $b = 6$ , and $c = 2$ which sum to $3+6+2 = 11$

@Jeffrey Wong . Take a look. :P Such solution, much long. Wow.