Question

不定积分一个问题，谢谢

不定积分（1+(sin 2x)^2)/(1+(cos 2x)^2) dx

Answer 1

∫ (1 + sin²2x)/(1 + cos²2x) dx
= ∫ [1 + (1 - cos4x)/2]/[1 + (1 + cos4x)/2] dx
= ∫ (3 - cos4x)/(3 + cos4x) dx
= ∫ [6 - (3 + cos4x)]/(3 + cos4x) dx
= 6∫ dx/(3 + cos4x) - ∫ dx
= (3/2)∫ d(4x)/(3 + cos4x) - x
= (3/2)∫ dθ/(3 + cosθ) - x，θ = 4x
= (3/2)M - x
M = ∫ dθ/(3 + cosθ)，令z = tan(θ/2)，dθ = 2dz/(1 + z²)，cosθ = (1 - z²)/(1 + z²)
= ∫ 1/[3 + (1 - z²)/(1 + z²)] * 2/(1 + z²) dz
= 2∫ 1/(3 + 3z² + 1 - z²) dz
= ∫ 1/(2 + z²) dz
= (1/√2)arctan(z/√2) + C'
= (1/√2)arctan[(1/√2)tan(θ/2)] + C'
= (1/√2)arctan[(1/√2)tan(2x)] + C'

所以∫ (1 + sin²2x)/(1 + cos²2x) dx = [3/(2√2)]arctan[(1/√2)tan(2x)] - x + C