# Evaluate the integrals \int_0^{\frac{\pi}{4}}\cos^2(4t-\frac{\pi}{4})dtZS

Evaluate the integrals
${\int }_{0}^{\frac{\pi }{4}}{\mathrm{cos}}^{2}\left(4t-\frac{\pi }{4}\right)dt$
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Given integral is ${\int }_{0}^{\frac{\pi }{4}}{\mathrm{cos}}^{2}\left(4t-\frac{\pi }{4}\right)dt$
Consider $u=4t-\frac{\pi }{4}$
Consider $u=4t-\frac{\pi }{4}$
Then,
$\frac{du}{dt}=4$
$dt=\frac{du}{4}$
Substitute $u=4t-\frac{\pi }{4}$ and $dt=\frac{du}{4}$ in ${\int }_{0}^{\frac{\pi }{4}}{\mathrm{cos}}^{2}\left(4t-\frac{\pi }{4}\right)dt$
$=\frac{1}{8}{\left[u+\frac{\mathrm{sin}\left(2u\right)}{2}\right]}_{-\frac{\pi }{4}}^{\frac{3\pi }{4}}$
$=\frac{1}{8}\left[\frac{3\pi }{4}+\frac{\pi }{4}+\frac{1}{2}\left(\mathrm{sin}\frac{3\pi }{2}-\mathrm{sin}\left(-\frac{\pi }{2}\right)\right)\right]$
$=\frac{\pi }{8}$