Answer
$$\int^{\pi/4}_0(1-\sin2t)^{3/2}\cos2tdt=\frac{1}{5}$$
Work Step by Step
$$A=\int^{\pi/4}_0(1-\sin2t)^{3/2}\cos2tdt$$
We set $u=1-\sin2t$, which means $$du=-2\cos2tdt$$ $$\cos2tdt=-\frac{1}{2}du$$
For $t=\pi/4$, we have $u=1-\sin\pi/2=1-1=0$
For $t=0$, we have $u=1-\sin0=1-0=1$
Therefore, $$A=-\frac{1}{2}\int^0_1u^{3/2}du$$ $$A=-\frac{1}{2}\times\frac{2u^{5/2}}{5}\Big]^0_1=-\frac{1}{5}u^{5/2}\Big]^0_1$$ $$A=-\frac{1}{5}(0-1)$$ $$A=\frac{1}{5}$$
