Question

integrate cos^4(t) dt

Answer 1

Using the identity \(\large \cos^2\alpha = \frac{1}{2}(1+\cos 2\alpha)\), Note that
\[\large \begin{aligned}\cos^4t &= (\cos^2t)^2\\ &= \left(\frac{1}{2}(1+\cos(2t)\right)^2\\ &= \frac{1}{4}(1+2\cos(2t)+\cos^2(2t))\\ &=\frac{1}{4}(1+2\cos(2t) +\frac{1}{2}(1+\cos(4t)))\\ &= \frac{1}{8}(3+4\cos(2t)+\cos(4t)) \end{aligned}\] Therefore
\[\large \int \cos^4t\,dt = \frac{1}{8}\int 3+4\cos(2t)+\cos(4t)\,dt\] Can you take things from here? :-)

Answer 2

sure! thanks for the help. ^_^