Integrating $\int_0^{\pi/2} \cos^a(x) \cos(bx) \ dx$

Please help me in this integral :

$$\int_0^{\pi/2} \cos^a(x) \cos(bx) \ dx \quad \text{if}\; b>a>-1$$

Please help me I used everything and can't evaluate it.

$$ \begin{align} &\int_0^{\frac\pi2}\cos^a(x)\cos(bx)\,\mathrm{d}x\\ &=\mathrm{Re}\left(2^{-a}\int_0^{\frac\pi2}\left(e^{ix}+e^{-ix}\right)^ae^{ibx}\,\mathrm{d}x\right)\tag{1}\\ &=\mathrm{Re}\left(2^{-a}\int_0^{\frac\pi2}\left(1+e^{-2ix}\right)^ae^{i(a+b)x}\,\mathrm{d}x\right)\tag{2}\\ &=\mathrm{Re}\left(2^{-a}\int_0^{\frac\pi2}\left(1+e^{-2ix}\right)^ae^{i(a+b+2)x}\,\frac{i}{2}\mathrm{d}e^{-2ix}\right)\tag{3}\\ &=\mathrm{Re}\left(2^{-a}\int_1^{-1}\left(1+u\right)^au^{-(a+b)/2-1}\,\frac{i}{2}\mathrm{d}u\right)\tag{4}\\ &=\mathrm{Re}\left(2^{-a}\int_0^{-1}\left(1+u\right)^au^{-(a+b)/2-1}\,\frac{i}{2}\mathrm{d}u\right)\tag{5}\\ &=\mathrm{Re}\left(2^{-a}\int_0^1\left(1-u\right)^au^{-(a+b)/2-1}e^{i\pi((a+b)/2+1)}\,\frac1{2i}\mathrm{d}u\right)\tag{6}\\ &=\mathrm{Re}\left(2^{-a-1}e^{i\pi((a+b+1)/2)}\int_0^1\left(1-u\right)^au^{-(a+b)/2-1}\,\mathrm{d}u\right)\tag{7}\\ &=2^{-a-1}\cos(\pi(a+b+1)/2)\int_0^1\left(1-u\right)^au^{-(a+b)/2-1}\,\mathrm{d}u\tag{8}\\ &=2^{-a-1}\sin(-\pi(a+b)/2)\frac{\Gamma(a+1)\Gamma(-(a+b)/2)}{\Gamma(1+(a-b)/2)}\tag{9}\\ &=\frac{\pi\,2^{-a-1}\,\Gamma(a+1)}{\Gamma(1+(a-b)/2)\,\Gamma(1+(a+b)/2)}\tag{10} \end{align} $$ Explanation:
$(1)$ write trigonometric functions as exponentials
$(2)$ pull $e^{iax}$ out of the parentheses
$(3)$ $\mathrm{d}x=e^{i2x}\frac i2\,\mathrm{d}e^{-2ix}$
$(4)$ $u=e^{-2ix}$ ($u$ travels clockwise along the unit circle from $1$ to $-i$ to $-1$
$\hphantom{(4)\ u=e^{-2ix}(}$ then Cauchy's theorem lets us deform the countour to
$\hphantom{(4)\ u=e^{-2ix}(}$ just under the real axis from $1$ to $0$ to $-1$)
$(5)$ the integral from $1$ to $0$ is pure imaginary, so we can drop it
$(6)$ $u\mapsto-u$ and since $u$ passed below $0$ we use $e^{-i\pi}$ for $-1$
$(7)$ pull the imaginary character out front
$(8)$ take the real part
$(9)$ use the integral for the Beta function
$(10)$ use the reflection formula $\Gamma(x)\Gamma(1-x)=\frac\pi{\sin(\pi x)}$ with $x=-(a+b)/2$