Answer
$$\int^{\pi/4}_{0}\cos^2\Big(4t-\frac{\pi}{4}\Big)dt=\frac{\pi}{8}$$
Work Step by Step
$$A=\int^{\pi/4}_{0}\cos^2\Big(4t-\frac{\pi}{4}\Big)dt$$
First, we set $u=4t-\frac{\pi}{4}$, which means $$du=4dt$$ $$dt=\frac{1}{4}du$$
For $t=0$, $u=-\pi/4$ and for $t=\pi/4$, $u=\pi-\pi/4=3\pi/4$
Therefore, $$A=\frac{1}{4}\int^{3\pi/4}_{-\pi/4}\cos^2udu=\frac{1}{4}B$$
We now consider B.
Recall the identity: $$\cos^2a=\frac{1+\cos2a}{2}$$
Apply the identity here, we have $$B=\int^{3\pi/4}_{-\pi/4}\frac{1+\cos2u}{2}du$$ $$B=\int^{3\pi/4}_{-\pi/4}\frac{1}{2}du+\frac{1}{2}\int^{3\pi/4}_{-\pi/4}\cos2udu$$ $$B=\frac{u}{2}\Big]^{3\pi/4}_{-\pi/4}+\Big(\frac{1}{2}\times\frac{1}{2}\sin2u\Big)\Big]^{3\pi/4}_{-\pi/4}$$ $$B=\frac{1}{2}\Big(\frac{3\pi}{4}-\Big(-\frac{\pi}{4}\Big)\Big)+\frac{1}{4}\sin2u\Big]^{3\pi/4}_{-\pi/4}$$ $$B=\frac{1}{2}(\pi)+\frac{1}{4}(\sin\frac{3\pi}{2}-\sin(-\frac{\pi}{2}))$$ $$B=\frac{\pi}{2}+\frac{1}{4}(-1-(-1))=\frac{\pi}{2}+\frac{1}{4}(0)$$ $$B=\frac{\pi}{2}$$
Therefore, $$A=\frac{1}{4}\times\frac{\pi}{2}=\frac{\pi}{8}$$
