微积分题求解答:1、∫[(tanθ-1)^2]dθ= 2、∫dθ/(1+sinθ)= 3、∫ln(x+1)dx=
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/16 09:34:53
微积分题求解答:1、∫[(tanθ-1)^2]dθ= 2、∫dθ/(1+sinθ)= 3、∫ln(x+1)dx=
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
=∫(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
微积分题求解答:1、∫[(tanθ-1)^2]dθ= 2、∫dθ/(1+sinθ)= 3、∫ln(x+1)dx=
求定积分:d/dx*[∫ (1到2)sin(x^2)dx]=
d/dx[∫(下限2 上限8)sin(ln x^2)dx]=
微积分 ∫ 1/(x ln^2 x )dx
这个∫1/ sin x dx 怎么等于ln (tan x/2) + ln c 我头转不过来了
∫ln(x+1)-lnx/x(x+1) dx =∫(ln(x+1)-lnx)d(ln(x+1)-lnx) =-1/2(l
∫f(x)dx=ln[sin(3x+1)]+C.求f(x)
定积分d/dx*[∫ (1到2)sin x^2dx]=
微积分求解:∫tan^2 (1/2 x) dx
积分练习题 ∫tan(x)dx 定积分在0到1/4π ∫(cos(x)ln(x)-sin(x)1/x)/ln^2 (x)
已知y=ln(x+√4+x²),求d²y/dx².求不定积分∫³√1+2lnx/
已知=ln[x+根号(x^2+1)],求二阶导数d^2y/dx^2