Question

1/(1+sin^2x)的不定积分如何求

Answer 1

∫ 1/(1+sin^2x)dx

= ∫ [1/cos^2x]/(1/cos^2x+tan^2x)dx

= ∫ [sec^2x]/(sec^2x + tan^2x)dx

= ∫ 1/(1 + 2tan^2x)dtanx

= 1/√2 *∫ 1/(1 + (√2tanx)^2)d(√2tanx)

= 1/√2 * arctan(√2tanx) + C