# How to integrate cos^2(2x)?

How to integrate ${\mathrm{cos}}^{2}\left(2x\right)$?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

2k1enyvp
I = integral
$I{\mathrm{cos}}^{2}\left(2x\right)dx=\left(\frac{1}{2}\right)I{\mathrm{cos}}^{2}\left(t\right)dt$
$2x=t,2dx=dt,dx=\frac{dt}{2}$
$I{\mathrm{cos}}^{2}\left(t\right)dt=I\mathrm{cos}t\mathrm{cos}tdt$
$u=\mathrm{cos}t,du=-\mathrm{sin}t$
$dv=\mathrm{cos}tdt,v=\mathrm{sin}t$
$I{\mathrm{cos}}^{2}\left(t\right)dt=\mathrm{sin}t\mathrm{cos}t+I{\mathrm{sin}}^{2}\left(t\right)dt$
$I{\mathrm{cos}}^{2}\left(t\right)dt=\frac{\mathrm{sin}\left(2t\right)}{2}+I\left(1-{\mathrm{cos}}^{2}\left(t\right)\right)dt$
$I{\mathrm{cos}}^{2}\left(t\right)dt=\frac{\mathrm{sin}\left(2t\right)}{2}+Idt-I{\mathrm{cos}}^{2}\left(t\right)dt$
$I{\mathrm{cos}}^{2}\left(t\right)dt+I{\mathrm{cos}}^{2}\left(t\right)dt=\frac{\mathrm{sin}\left(2t\right)}{2}$
$2I{\mathrm{cos}}^{2}\left(t\right)dt=\frac{\mathrm{sin}\left(2t\right)}{2}+t$
$I{\mathrm{cos}}^{2}\left(t\right)dt=\frac{\mathrm{sin}\left(2t\right)}{4}+\frac{t}{2}$
t=2x
$=\frac{\mathrm{sin}\left(4x\right)}{4}+x$