Question

(1-sinx平方)/(1+sinx平方)的不定积分

Answer 1

∫ (1 - sin²x)/(1 + sin²x) dx
= ∫ [1 - (1 - cos2x)/2]/[1 + (1 - cos2x)/2] dx
= ∫ (1 + cos2x)/(3 - cos2x) dx
= ∫ [4 - (3 - cos2x)]/(3 - cos2x) dx
= 4∫ dx/(3 - cos2x) - ∫ dx
= 2∫ du/(3 - cosu) - x，u = 2x
= 2∫ 2dz/(1 + z²) * 1/[3 - (1 - z²)/(1 + z²)] - x，z = tan(u/2)
= 2∫ dz/(1 + 2z²) - x
= (2/√2)∫ d(√2z)/[1 + (√2z)²] - x
= √2arctan(√2z) - x + C
= √2arctan[√2tan(u/2)] - x + C
= √2arctan(√2tanx) - x + C