Answer
$$\int^{\pi/6}_03\cos^53xdx=\frac{8}{15}$$
Work Step by Step
$$A=\int^{\pi/6}_03\cos^53xdx$$ $$A=\int^{\pi/2}_0\Big(3\cos^53x\Big)\frac{1}{3}d(3x)$$ $$A=\int^{\pi/2}_0\cos^53xd(3x)$$
We set $a=3x$.
$$A=\int^{\pi/2}_0\cos^5ada$$
Following the integral form $\int\sin^mx\cos^nxdx$, we would apply Case 2 here, where $m=0$ and $n=5$.
$$A=\int^{\pi/2}_0(\cos^2a)^2\cos ada$$ $$A=\int^{\sin(\pi/2)}_{\sin0}(1-\sin^2a)^2d(\sin a)$$
We set $u=\sin a$.
Limits of integration: $\sin(\pi/2)=1$ and $\sin0=0$
$$A=\int^1_0(1-u^2)^2du$$ $$A=\int^1_0(1-2u^2+u^4)du$$ $$A=\Big(u-\frac{2u^3}{3}+\frac{u^5}{5}\Big)\Big]^1_0$$ $$A=\Big(1-\frac{2}{3}+\frac{1}{5}\Big)-0$$ $$A=\frac{8}{15}$$
