Question

∫sin^2x/(1+sin^2x )dx求解，在线等

Answer 1

解：
原式=∫[1-1/(2sin²x+cos²x)]dx
=∫dx - ∫[sec²x/(2tan²x+1)]dx
=∫dx - ∫1/[(√2tanx)²+1)]d(tanx)
=∫dx - √2/2* ∫1/[(√2tanx)²+1)]d(√2tanx)
=x - ∫dx - √2/2*arctan(√2tanx)+c

Answer 2

原式=∫[1-1/(2sin²x+cos²x)]dx
=∫dx - ∫[sec²x/(2tan²x+1)]dx
=∫dx - ∫1/[(√2tanx)²+1)]d(tanx)
=∫dx - √2/2* ∫1/[(√2tanx)²+1)]d(√2tanx)
=x-√2/2*arctan(√2tanx)+c

Answer 3

∫sin^2x/(1+sin^2x )dx
=∫（(1+sin^2x )-1）/(1+sin^2x )
=x-1/(1+sin^2x )
=x-1/(cos^2x +2sin^2x )√
=x-csc^2x /(cot^2x +2)
=x+（1/√2）argtan（cot^2x /√2）