550 Followers Problem

$\begin{aligned} \int_0^\pi \frac{\sin^4x \sqrt{1-\cos x}}{(1+\cos{x})^2} dx & = \int_0^\infty \frac{\left(\frac{2t}{1+t^2} \right) ^4 \sqrt{1-\frac{1-t^2}{1+t^2}}}{\left(1+\frac{1-t^2}{1+t^2}\right)^2}\dot{} \frac{2}{1+t^2} dt \quad \quad \small \color{#3D99F6} {\text{Let } t = \tan \left(\frac{x}{2}\right) \quad \Rightarrow dx = \frac{2}{1+t^2} dt} \\ & = \int_0^\infty \frac{\left(\frac{2t}{1+t^2} \right) ^4 \sqrt{\frac{2t^2}{1+t^2}}}{\left(\frac{2}{1+t^2}\right)^2}\dot{} \frac{2}{1+t^2} dt \\ & = \int_0^\infty \frac{2^{\frac{7}{2}}t^5}{(1+t^2)^{\frac{7}{2}}} dt \quad \quad \small \color{#3D99F6} {\text{Let } u^2 = 1+t^2 \quad \Rightarrow 2u \space du = 2t \space dt \quad \Rightarrow u \space du = t \space dt} \\ & = 8\sqrt{2} \int_1^\infty \frac{u(u^2-1)^2}{u^7} du \\ & = 8\sqrt{2} \int_1^\infty \frac{u^5-2u^3+u}{u^7} du \\ & = 8\sqrt{2} \int_1^\infty \left( u^{-2}-2u^{-4}+u^{-6} \right) du \\ & = 8\sqrt{2} \left[ - \frac{1}{u} + \frac{2}{3u^3} - \frac{1}{5u^5} \right]_1^\infty \\ & = 8\sqrt{2} \left( 1 - \frac{2}{3} + \frac{1}{5} \right) \\ & = \frac{64\sqrt{2}}{15} \end{aligned}$

$\Rightarrow A+B+C = 64+2+15 = \boxed{81}$

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Solution as suggested by Pi Han Goh

$\begin{aligned} \int_0^\pi \frac{\sin^4x \sqrt{1-\cos x}}{(1+\cos{x})^2} dx & = \int_0^\pi \frac{2^4\sin^4 \frac{x}{2} \cos^4\frac{x}{2}\sqrt{1-1+2\sin^2 \frac{x}{2}}}{\left(1+2\cos^2 \frac{x}{2}-1\right)^2} dx \\ & = \int_0^\pi \frac{2^4\sqrt{2} \sin^5 \frac{x}{2} \cos^4\frac{x}{2}}{2^2 \cos^4 \frac{x}{2}} dx \\ & = 4\sqrt{2} \int_0^\pi \sin^5 \frac{x}{2} \space dx \quad \quad \small \color{#3D99F6}{\text{Let } u = \frac{x}{2} \quad \Rightarrow dx = 2 du} \\ & = 8\sqrt{2} \int_0^\frac{\pi}{2} \sin^5 u \space du \quad \quad \small \color{#3D99F6}{\text{Let } v = \cos{u} \quad \Rightarrow \sin^2 = 1-v^2 \quad \Rightarrow du = -\sin u \space dv} \\ & = 8\sqrt{2} \int_0^1 \left(1-v^2\right)^2 \space dv \\ & = 8\sqrt{2} \int_0^1 \left(1- 2 v^2 + v^4 \right) \space dv \\ \\ & = 8\sqrt{2} \left[v - \frac{2v^3}{3} + \frac{v^5}{5} \right]_0^1 \\ & = 8\sqrt{2} \left( 1 - \frac{2}{3} + \frac{1}{5} \right) \\ & = \frac{64\sqrt{2}}{15} \end{aligned}$

$\Rightarrow A+B+C = 64+2+15 = \boxed{81}$

Rewriting sinx^4 as (1-cosx)^2(1+cosx)^2 you come to the integral of (1-cosx)^5/2 from 0 to pi that gives the result