A calculus problem by Jose Sacramento

$\begin{aligned} I & = \int_0^\pi \frac {dx}{(5-3\cos x)^3} \quad \quad \small \color{#3D99F6}{\text{Let }t=\tan \frac x2, \ dx = \frac {2 \ dt}{1+t^2}, \ \cos x = \frac {1-t^2}{1+t^2}} \\ & = \int_0^\infty \frac {2 \ dt}{(1+t^2) \left( 5-3\left(\frac {1-t^2}{1+t^2}\right) \right)^3} \\ & = \int_0^\infty \frac {2 (1+t^2)^2}{ \left(2+ 8 t^2 \right)^3}dt \\ & = \int_0^\infty \frac {(1+t^2)^2}{4 \left(1+ 4 t^2 \right)^3}dt \quad \quad \small \color{#3D99F6}{\text{By partial fractions}} \\ & = \frac 1{64} \int_0^\infty \frac 1{1+4t^2} dx + \frac 3{32} \int_0^\infty \frac 1{(1+4t^2)^2} dx + \frac 9{64} \int_0^\infty \frac 1{(1+4t^2)^3} dx \end{aligned}$

Now, let $u = 4 t^2 \implies t = \dfrac {u^\frac 12}2 \implies dt = \dfrac {u^{-\frac 12}}4 du$ . Then we have:

$\begin{aligned} I & = \frac 1{64} \int_0^\infty \frac 1{1+4t^2} dx + \frac 3{32} \int_0^\infty \frac 1{(1+4t^2)^2} dx + \frac 9{64} \int_0^\infty \frac 1{(1+4t^2)^3} dx \\ & = \frac 1{256} \int_0^\infty \frac {u^{-\frac 12}}{1+u} du + \frac 3{128} \int_0^\infty \frac {u^{-\frac 12}}{(1+u)^2} du + \frac 9{256} \int_0^\infty \frac {u^{-\frac 12}}{(1+u)^3} du \\ & = \frac 1{256} \text B \left(\frac 12, \frac 12 \right) + \frac 3{128} \text B \left(\frac 12, \frac 32 \right) + \frac 9{256} \text B \left(\frac 12, \frac 52 \right) \ \ \ \ \ \ \ \ \small \color{#3D99F6}{\text{B}(\cdot) \text{ is beta function}} \\ & = \frac 1{256} \cdot \frac {\Gamma \left(\frac 12 \right)\Gamma \left(\frac 12 \right)}{\Gamma \left(1 \right)} + \frac 3{128} \cdot \frac {\Gamma \left(\frac 12 \right)\Gamma \left(\frac 32 \right)}{\Gamma \left(2 \right)} + \frac 9{256} \cdot \frac {\Gamma \left(\frac 12 \right)\Gamma \left(\frac 52 \right)}{\Gamma \left(3 \right)} \ \ \ \ \ \ \ \ \small \color{#3D99F6}{\Gamma (\cdot) \text{ is gamma function}} \\ & = \frac 1{256} \cdot \frac {\sqrt \pi \cdot \sqrt \pi}{0!} + \frac 3{128} \cdot \frac {\sqrt \pi \cdot \frac 12 \cdot \sqrt \pi}{1!} + \frac 9{256} \cdot \frac {\sqrt \pi \cdot \frac 12 \cdot \frac 32 \cdot \sqrt \pi}{2!} \\ & = \left(\frac 1{256} + \frac 3{256} + \frac {27}{2048} \right) \pi \\ & = \frac {59 \pi}{2048} \end{aligned}$

$\implies A+B = 59+2048 = \boxed{2107}$

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References:

• Partial fractions
• Beta function
• Gamma function