Terrible Integrals!

Relevant wiki: Beta Function

$\begin{aligned} I & = \int_0^\frac \pi 2 \sin^5 x \cos^3 x \ dx \\ & = \frac 12 \color{#3D99F6}{B(3,2)} & \small \color{#3D99F6}{\text{where }B(m,n) \text{ is beta function.}} \\ & = \frac 12 \cdot \frac {2! \cdot 1!}{4!} \\ & = \frac 1{24} \end{aligned}$

$\implies a+b = 1+24 = \boxed{25}$

$\displaystyle I = \int \sin^5 x\cos^3 x ~ dx\\ \displaystyle \boxed{y = \sin x \implies \dfrac{dy}{dx} = \cos x \\ \displaystyle \implies \cos x ~ dx = dy} \\ \displaystyle I = \int \sin^5 x \left(\cos^2 x\right)\cos x ~ dx \\ \displaystyle = \int sin^5 x \left(1 - \sin^2 x \right)\cos x ~ dx \\ \displaystyle = \int y^5 \left(1 - y^2\right) ~ dy \\ \displaystyle = \int y^5 - y^7 ~ dy \\ \displaystyle = \int y^5 dy - \int y^7 dy \\ \displaystyle = \dfrac{y^6}{6} - \dfrac{y^8}{8} \\ \displaystyle = \dfrac{\sin^6(x)}{6} - \dfrac{\sin^8(x)}{8} \displaystyle \therefore \int_0^{\dfrac{\pi}{2}}\sin^5 x\cos^3 x ~ dx = \dfrac{\sin^6(\dfrac{\pi}{2})}{6} - \dfrac{\sin^8(\dfrac{\pi}{2})}{8} - (\dfrac{\sin^6(0)}{6} - \dfrac{\sin^8(0)}{8}) \\ \displaystyle \boxed{= \dfrac{1}{6} - \dfrac{1}{8} = \dfrac{1}{24} \\ \displaystyle \therefore a+b = 1 + 24 = 25}$

Put $y = sin^6 x$ and then the integral reduces to evaluating $\frac{1}{6}$ (y- $\frac{3}{4}$ y^(1/3)) from 0 to 1 which equals $\frac{1}{24}$ hence final answer is $1 + 24$ = $\boxed{25}$

Just 2 words , $Walli's\ Formula$ . $$ why it's level 4 ?

Ha ha easy peasy .

Let sinx = t

Integration from 0 to 1 of (t^5 - t^7)