삼각함수 적분 ∫ (sin x)^4 dx

[수학짱]

(sin x)^4 = (sin^2 x)^2 = {(1-cos 2x)/2}^2

= (1 - 2cos 2x + cos^2 2x)/4 

=  {1 - 2cos 2x + (1+cos 4x)/2}/4

= (3 - 4cos 2x + cos 4x)/8

이므로

∫ (sin x)^4 dx =  { 3x - 2sin 2x + (1/4) sin 4x} /8 + C

참고  http://cafe.naver.com/calculus/1651