It would be easier to generalize this problem -2

$\begin{aligned} \int_0^{\pi/2}\cos^9x\,\mathrm dx&=\int_0^{\pi/2}(1-\sin^2 x)^4\cos x\,\mathrm dx\\ &=\int_0^{1}(1-t^2)^4\,\mathrm dt\\ &=\int_0^{1}(1-4t^2+6t^4-4t^6+t^8)\,\mathrm dt \end{aligned}$

$\text{Game Over!}$

\begin{aligned} I & = \int_0^\frac \pi 2 \cos^9 x \ dx \\ & = \int_0^\frac \pi 2 \cos^\color{#3D99F6}{9} x \sin^\color{#3D99F6}{0} x \ dx \\ & = \frac 12 \color{#3D99F6}{B} \left( \frac {\color{#3D99F6}{9}+1}2, \frac {\color{#3D99F6}{0}+1}2 \right) & \small \color{#3D99F6}{B(m,n) \text{ is beta function}} \\ & = \frac 12 B \left( 5, \frac 12 \right) \\ & = \frac {\Gamma (5) \Gamma \left(\frac 12 \right)}{ 2 \Gamma \left(5\frac 12 \right)} & \small \color{#3D99F6}{\Gamma (x) \text{ is gamma function}} \\ & = \frac {4! \sqrt \pi}{2 \cdot \frac 92 \cdot \frac 72 \cdot \frac 52 \cdot \frac 32 \cdot \frac 12 \sqrt \pi } \\ & = \frac {128}{315} \end{aligned}

$\implies a-b = 128-315 = \boxed{-187}$

We have Formulas for cosines, called Wallis cosine formulas.

$\large\int_{0}^{\pi/2} cos^{2n+1} x dx= \dfrac{2.4.6...(2n)}{1.3.5...(2n+1)}= \dfrac{2^{n} n!}{2^{n}(n-1/2)!\sqrt{\pi}}=\frac{128}{315} for \ n= 4 \implies a-b=128-315=-187$

I derived the reduction formula using integration by parts

We consider the general integral

I = cos^n(x) dx

Now separating functions and applying integration by parts using first function as (cosx)^(n-1) and second function as cosx.

We get the general reduction formula

I(n)/I(n-2) =n-1/n

Now rest is trivial its a telescoping product which can be evaluated easily!

Beta Function kills it........!

We can use Wallis formula in solving this problem..

I did it using gamma function .