Answer
$\int_{0}^{\pi} cos^4(2t)dt = \frac{3\pi}{8}$
Work Step by Step
$\int_{0}^{\pi} cos^4(2t)dt=$
Break up the $cos^4(2t)$:
$\int_{0}^{\pi} cos^2(2t)cos^2(2t)dt=$
Use the power reduction formula:
$cos^2(x)=\frac{1}{2}(1-cos(2x))$
$\int_{0}^{\pi}\frac{1}{2}(1-cos(2*2t))\frac{1}{2}(1-cos(2*2t))dt$
Clean up the integral:
$\frac{1}{4 }\int_{0}^{\pi}1-2cos(4t)+cos^2(4t)dt$
Do the integrals independently:
$\frac{1}{4 }[ \int_{0}^{\pi}1dt-2\int_{0}^{\pi}cos(4t)dt+\int_{0}^{\pi}cos^2(4t)dt]$
Solve the integrals:
$\frac{1}{4 }[(\pi-0)-2(\frac{sin(4\pi)}{4}-\frac{sin(0)}{4})+\int_{0}^{\pi}cos^2(4t)dt]$
$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$
Solving this integral:
$ \int_{0}^{\pi}cos^2(4t)dt $
Apply the power reduction formula:
$\frac{1}{2}\int_{0}^{\pi}1-cos(2*4t)dt$
Apply the same principle as before:
$\frac{1}{2}[(\pi-0)-(\frac{sin(8\pi)}{8}-\frac{sin(0)}{8})]$
$= \frac{\pi}{2}$
$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$
Back to the original problem:
$\frac{1}{4 }[(\pi-0)-2(\frac{sin(4\pi)}{4}-\frac{sin(0)}{4})+(\frac{\pi}{2})]=$
$\frac{\pi}{4}+\frac{\pi}{8}=\frac{3\pi}{8}$
