∫sin^4θdθ
=∫(sin^2θ)^2dθ
=∫[(1-cos2θ)/2]^2dθ
=1/4∫[1-2cos2θ+cos^2(2θ)]dθ
=1/4θ-1/4sin2θ+1/4∫[(1+cos4θ)/2]dθ
=1/4θ-1/4sin2θ+1/8θ+1/8∫cos4θdθ
=3/8θ-1/4sin2θ+1/32sin4θ+C
∫sin⁴θdθ
=∫sin²θ(1-cos²θ)dθ
=∫sin²θ-sin²θcos²θdθ
=∫[½(1-cos2θ)-¼sin²2θ]dθ
=∫[½(1-cos2θ)-⅛(1-cos4θ)]dθ
=∫(⅛cos4θ-½cos2θ+⅜)dθ
=(1/32)∫cos4θd(4θ)- ¼∫cos2θd(2θ)+∫⅜dθ
=(1/32)sin(4θ)-¼sin(2θ)+⅜θ +C
=[sin(4θ)-8sin(2θ)+12]/32 +C
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024