早教吧作业答案频道 -->数学-->
求定积分∫(0~a)1/[x+√(a^2-x^2)]dx
题目详情
求定积分∫ (0~a) 1/[x+√(a^2-x^2)] dx
▼优质解答
答案和解析
令x = asinθ,dx = acosθdθ
原式= ∫(0→π/2) (acosθ)/(asinθ + acosθ) dθ
= (1/2)∫(0→π/2) 2cosθ/(sinθ + cosθ) dθ
= (1/2)∫(0→π/2) [(sinθ + cosθ) - (sinθ - cosθ)]/(sinθ + cosθ) dθ
= (1/2)∫(0→π/2) dθ - (1/2)∫(0→π/2) (sinθ - cosθ)/(sinθ + cosθ) dθ
= (1/2)(π/2) - (1/2)∫(0→π/2) - d(cosθ + sinθ)/(sinθ + cosθ) dθ
= π/4 + (1/2)ln(sinθ + cosθ) |(0→π/2)
= π/4 + (1/2)[ln(1 + 0) - ln(0 + 1)]
= π/4
原式= ∫(0→π/2) (acosθ)/(asinθ + acosθ) dθ
= (1/2)∫(0→π/2) 2cosθ/(sinθ + cosθ) dθ
= (1/2)∫(0→π/2) [(sinθ + cosθ) - (sinθ - cosθ)]/(sinθ + cosθ) dθ
= (1/2)∫(0→π/2) dθ - (1/2)∫(0→π/2) (sinθ - cosθ)/(sinθ + cosθ) dθ
= (1/2)(π/2) - (1/2)∫(0→π/2) - d(cosθ + sinθ)/(sinθ + cosθ) dθ
= π/4 + (1/2)ln(sinθ + cosθ) |(0→π/2)
= π/4 + (1/2)[ln(1 + 0) - ln(0 + 1)]
= π/4
看了 求定积分∫(0~a)1/[x...的网友还看了以下:
求下列不定积分.求详解(1)∫[(cos2x)/(cosx﹢sinx)]dx;(2)∫cot²xdx 2020-03-31 …
陈文灯《复习指南》中定积分一道计算题·设函数f(x),g(x)满足f'(x)=g(x),g'(x) 2020-04-26 …
求下列不定积分2“——”前为题目,后为该题答案,我没有凑出来答案数.1.∫xe^(3x^2)dx— 2020-05-20 …
已知复合函数f(e^x)=e^x+x求不定积分∫f(x)dx求不定积分∫√(x-1)^3/xdx第 2020-06-03 …
考试中急.不定积分Jx方inxdx.J1/((1+x方)x方)dx定积分J上4下01/(2+... 2020-06-22 …
∫(0,1)√(2x-x^2)dx=?定积分的应用, 2020-07-11 …
f(x)连续,且定积分0到3f(x)dx=3,定积分0到4f(x)dx=7,求(1)定积分3到4f 2020-07-25 …
求不定积分∫1/[1+√(2x)]dx第二类换元法∫1/[1+√(1-x^2)]dx第二类换元法∫ 2020-08-01 …
求不定积分有人说dx可以不要,F'(x)=f(x),即f(x)的不定积分是F(x)+c,为什么加d 2020-08-03 …
求不定积分∫x^2+1/(x^2-1)(x+1)dx答案是1/2ln(x^2-1)+1/求不定积分∫ 2020-10-31 …