∫sinx+cosxdx

储强惠3505求∫cosx/(sinx+cosx)dx的不定积分 -
鲍扶苏18922047707 ______ ∫cosx/(sinx+cosx) dx= (1/2)∫[(cosx+sinx)+(cosx-sinx)]/(sinx+cos)] dx= (1/2)∫ dx + (1/2)∫(cosx-sinx)/(sinx+cosx) dx= x/2 + (1/2)...

储强惠3505求∫1/1+cosxdx的不定积分? -
鲍扶苏18922047707 ______ ∫ 1/(1+cosx) dx=(1/2)∫ 1/cos²(x/神唯2) dx=∫ sec²(x/2) d(x/2)=tan(x/2) + C希望可以帮到你,孝搜如有疑问请追问,如满意请点“选巧瞎历为满意答案”.

储强惠3505求不定积分∫(x²+cosx)dx -
鲍扶苏18922047707 ______ ∫(x²+cosx)dx=∫x²dx+∫cosxdx=x³/3+sinx+C

储强惠3505高数∫(cosX+sinX)dx -
鲍扶苏18922047707 ______[答案] 原式=∫cosXdx+∫sinXdx =sinX-cosX+C 因为cosX积分等于sinX,sinX积分等于-cosX,这是公式,记着就行.

储强惠3505∫x乘以cosxdx -
鲍扶苏18922047707 ______ ∫x乘以cosxdx =∫xdsinx=xsinx-∫sinxdx=xsinx+cosx+c(c为常数)

储强惠3505求不定积分∫[sinxcosx/(sinx+cosx)]dx -
鲍扶苏18922047707 ______ ∫[sinxcosx/(sinx+cosx)]dx=-1/4∫[dcos2x/(sinx+cosx)]=-1/4cos2x/(sinx+cosx)-1/4/∫[cos2x*(cosx-sinx)/(sinx+cosx)^2dx=-1/4cos2x/(sinx+cosx)-1/4∫[(cosx-sinx)^2/(sinx+cosx)]dx=-1/4cos2x/(sinx+cosx)-1/4∫[(1-2sinx*cosx)/(sinx+cosx)]dx=-1/4cos2x/(sinx...

储强惠3505求∫x/2+cosxdx 和∫1/x√x dx -
鲍扶苏18922047707 ______ ∫f(sinx)cosxdx=∫f(sinx)dsinx因为∫f(x)dx=1/(1+x^2) +c 所以∫f(sinx)dsinx = 1/[1+(sinx)^2] +c 那么∫f(sinx)cosxdx =1/[1+(sinx)^2] +c

储强惠3505求n∫【n,0】|sinx+cosx|dx的值 -
鲍扶苏18922047707 ______[答案] n∫{n,0}|sinx+cosx|dx=n*√2∫{n,0}|sin(x+л/4)|dx∫{0,л}|sin(x+л/4)|dx=∫{0,3л/4}sin(x+л/4)dx-∫{3л/4,л}sin(x+л/4)dx=(cos0-cosл)=2;∫{0,θ}|sin(x+л/4)|dx=∫{0,3л/4}sin(x+л/4)dx-∫{3л/4,θ...

储强惠3505计算定积分∫(π/2~0) x(sinx+cosx) dx -
鲍扶苏18922047707 ______ ∫x(sinx+cosx)dx =∫xsinxdx+∫xcosxdx =-∫xd(cosx)+∫xcosxdx =-xcosx+∫cosxdx+∫xd(sinx) =-xcosx+sinx+xsinx-∫sinxdx =-xcosx+sinx+xsinx+cosx ∫(π/2~0) x(sinx+cosx)=(-xcosx+sinx+xsinx+cosx )|(π/2~0) =(1+π/2)-(1)=π/2

储强惠3505∫(sin^3x) /1+cosxdx不定积分 -
鲍扶苏18922047707 ______[答案] ∫(sin^3x) /(1+cosx)dx =∫sinx(1-cos^2x) /(1+cosx)dx =∫(1+cosx)sinxdx =∫sinxdx+1/2*∫sin2xdx =-cosx-1/4*cos2x+C