Question

碰到一个积分的题目，看着很简单，但是却不会弄了

Answer 1

∫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

Answer 2

∫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