Question

微积分题求解答：1、∫[(tanθ-1)^2]dθ= 2、∫dθ/(1+sinθ)= 3、∫ln(x+1)dx=

Answer 1

1、∫(tanθ-1)²dθ
=∫(tan²θ-2tanθ+1)dθ
=∫(sec²θ-2tanθ)dθ
=tanθ+2ln|cosθ|+C

2、∫dθ/(1+sinθ)
=∫1/(1+2sin(θ/2)cos(θ/2)) dθ
分子分母同除以cos²(θ/2)，得
=∫sec²(θ/2)/(sec²(θ/2)+2tan(θ/2)) dθ
=2∫sec²(θ/2)/(tan²(θ/2)+1+2tan(θ/2)) d(θ/2)
=2∫1/(tan(θ/2)+1)² d(tan(θ/2))
=-2/(tan(θ/2)+1)+C

3、∫ln(x+1)dx
=xln(x+1)-∫x/(x+1)dx
=xln(x+1)-∫(x+1-1)/(x+1)dx
=xln(x+1)-∫1dx+∫1/(x+1)dx
=xln(x+1)-x+ln(x+1)+C

Answer 2

这就是结果：