∫sin^4θdθ不定积分怎么求，具体解法。

Answer 1

降次
∫sin^4θdθ
=∫(sin²θ)²dθ
=∫(1-cos2θ)²/4dθ
=1/4∫(1-2cos2θ+cos²2θ)dθ
=1/4∫dθ-1/4∫cos2θd(2θ)+1/4∫(1+cos4θ)/2dθ
=1/4θ-1/4sin2θ+1/8∫dθ+1/32∫cos4θd(4θ)
=3/8θ-1/4sin2θ+1/32sin4θ+C

Answer 2

原式=1/4∫sin^4θd4θ
=-1/4cos4θ

Answer 3

∫sin^4θdθ
=∫[sin²θ]²dθ
=∫[1-cos²θ]²dθ
cos²θ=(1+cos2θ)/2
1-cos²θ=(1-cos2θ)/2
=1/4∫[(1-cos2θ)]²dθ
=1/4∫(1+cos²2θ-2cos2θ)dθ
=1/4θ+1/4∫cos²2θdθ-1/2∫cos2θdθ
=θ/4+1/4∫(1+cos4θ)/2dθ-1/4∫cos2θd2θ
=θ/4+1/8θ+1/32∫cos4θd4θ-1/4sin2θ
=3/8θ+1/32sin4θ-1/4sin2θ+C

Answer 4

(sinθ)^4 = [(1-cos2θ)/2]² = [1-2cos2θ+(cos2θ)²]/4
= 3/8 - cos2θ /2 + cos4θ /8
∫ (sinθ)^4 dθ = ∫ [3/8 - cos2θ /2 + cos4θ /8] dθ
= 3θ/8 - sin2θ /4 + sin4θ /32 + C