Party time

If one lets $x=\frac{3}{2} \tan\theta$ the integral simplifies to $\displaystyle \int_{0}^{\frac{\pi}{3}} \frac{3}{16} \times \frac{(\sin \theta)^3}{(\cos \theta)^2} d\theta =\displaystyle \int_{0}^{\frac{\pi}{3}} \frac{3}{16} \times \frac{(1-(\cos \theta)^2)(\sin\theta)}{(\cos \theta)^2}d\theta$ . The resulting integral can be split along the minus sign in the numerator and finally solved using a u-substitution, leading to $\frac{3}{32}$ with a sum of $35$ .

I used the substitution $u=4{ x }^{ 2 }+9$ , which gives the following integral:

$\displaystyle \frac { 1 }{ 32 } \displaystyle \int _{ 9 }^{ 36 }{ \frac { u-9 }{ { u }^{ \frac { 3 }{ 2 } } } du } =\frac { 1 }{ 32 } \displaystyle \int _{ 9 }^{ 36 }{ \left( { u }^{ \frac { -1 }{ 2 } }-9{ u }^{ \frac { -3 }{ 2 } } \right) du }$

Which evaluates to $\frac { 3 }{ 32 }$ with a sum of $35$ .

To clarify, the numerator of the fraction I created with $u$ comes from rearranging my substitution for $u$ .

Use integration by parts to get the integral. Substitute the two given values to it and simplify. The answer will be 3/32. However, 32 is not a prime number, because it can be factored by other numbers excluding 1.