Question

求∫1/(1+sin^2x)dx的积分,上下线为0到π/2

Answer 1

令tanx = t,x = arctant

则dx = dt/(1+t²)

1+sin²x = 1 + t²/(1+t²)

∫dx/(1+sin²x)

=∫dt/(1+2t²)

=1/√2 ∫d√2t/[1+(√2t)²]

=1/√2 arctan√2t + C

=1/√2 arctan(√2tanx) + C

x = π/2时,1/√2 arctan(√2tanx) = π/2√2

x = 0时,1/√2 arctan(√2tanx) = 0

原式 = π/2√2