Calculus: Early Transcendentals 8th Edition

$\int_{0}^{\pi} cos^4(2t)dt = \frac{3\pi}{8}$
$\int_{0}^{\pi} cos^4(2t)dt=$ 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: $\frac{1}{4 }[(\pi-0)-2(\frac{sin(4\pi)}{4}-\frac{sin(0)}{4})+(\frac{\pi}{2})]=$ $\frac{\pi}{4}+\frac{\pi}{8}=\frac{3\pi}{8}$