4∫(pi/2,0)(1-cos^4θ)dθ

Question

4∫(pi/2,0)(1-cos^4θ)dθ

Answer 1

原式=4∫[0,π/2][1-(cosθ)^2][1+(cosθ)^2]dθ
=4∫[0,π/2](1/4)[1-cos2θ)(3+cos2θ)dθ
=∫[0,π/2][3-2cos2θ-(cos2θ)^2]dθ
=[0,π/2][3θ-sin2θ]-(1/2)∫[0,π/2](1+cos4θ)dθ
=3π/2-0-(1/2)θ[0,π/2]-(1/8)∫[0,π/2]cos4θd(4θ)
=5π/4-(1/8)sin4θ[0,π/2]
=5π/4.