Question

求下图中的定积分，谢谢

Answer 1

∫(0->π) (sin2θ)^2 dθ
=(1/2) ∫(0->π) (1-cos4θ) dθ
=(1/2) [ θ - (1/4)sin4θ] |(0->π)
=π/2
√[ (1/π) .∫(0->π) (sin2θ)^2 dθ ]
=√(1/2)
=√2/2

追问

这个呢？

追答

∫(0->π/2) (sin4θ)^2 dθ
=(1/2)∫(0->π/2) (1-cos8θ) dθ
=(1/2)[θ-(1/8)sin8θ]|(0->π/2)
=π/4
/
[1/(π/2) ] .∫(0->π/2) (sin4θ)^2 dθ
=1/2

追问

是不是恒为1/2啊？？

追答

n=1,2,3,...
∫(0->2π/n) (sinnθ)^2 dθ

=(1/2) ∫(0->2π/n) (1-cos2nθ) dθ
=(1/2) [θ- (1/(2n))sin2nθ) ]|(0->2π/n)
=π/n
=>
[n/(2π)].∫(0->2π/n) (sinnθ)^2 dθ=1/2