方法一:运用倍角公式
(sint)^4
=(sin²t)²
=((1-cos2t)/2)²
=1/4-1/2*cos2t+1/4*cos²2t
=1/4-1/2*cos2t+1/8*(cos4t+1)
=3/8-1/2*cos2t+1/8*cos4t
所以
∫(sint)^4 dt (积分范围0→π/2)
=∫(3/8-1/2*cos2t+1/8*cos4t)dt(积分范围0→π/2)
=3t/8-1/4*sin2t+1/32*sin4t(积分范围0→π/2)
=3π/16
方法二:直接利用公式
∫(sint)^4 dt (积分范围0→π/2)
=3π/16
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因此
2∫[1-(sint)^4] dt (积分范围0→π/2)
=2t-2∫(sint)^4 dt (积分范围0→π/2)
=π-3π/8
=5π/8