How do you find the integral of #int [sin^2(pi x) cos^5(pi x)]dx#?
1 Answer
Oct 11, 2015
This belongs in your mathematical toolbox:
Explanation:
Integrate by substitution. Do this by pulling off one from the odd power, then convert the remaining even power to the other function. Integrate the resulting polynomial in
#= int [sin^2pix -2sin^4 pix+sin^6 pix]cos(pix)dx#
# = 1/pi int (u^2-2u^4+u^6)du#
# = 1/pi [(sin^3 pix)/3-(2sin^5 pix)/5 + (sin^7 pix)/7] +C#