Do any one have an idea how to calculate integral of (cos x)^2 ? Or is it even possible? I tried some substitutions and/or rules of trigonometry, like cosxcosx+sinxsinx=1, but it didn't help. Thank you!
an easy way to remember the solution to this common integral, when integrating over a whole period: cos^2 x + sin ^2 x =1 [tex] \int cos^2 x = \int sin^2 x [/tex] , at least when you integrate over a whole period [tex] \int cos^2 x + \int sin^2 x =[/tex] length of a period so the integral gives length of a period divided by 2
First use the half-angle formula to change the cos(x)^2 to (1+cos(2x))/2... This will allow you to break the integral into two seperate problems much easier to solve integral{ 1/2dx + integral{ cos(2x)dx Then you will have x/2 + (sin(2x)/2) + C
What the, that's not even correct. If you're gonna revive a 5-year old thread, at least make sure you don't have arithmetic errors.
http://www.5min.com/Video/An-Introduction-to-Integrals-with-Powers-of-Sine-and-Cosine-169056088 Why doesn't the student, after nearly 6 years of unsuccessfully attempting this crazy integral, try a visual aid?