d/dx∫sin(x-t)^2dt 积分上限x下限0
先求∫(0->x)sin(x-t)^2dt
=∫(0->x)(1-cos(2x-2t)/2 dt
=1/2∫(0->x)dt-1/2∫(0->x)cos(2x-2t)dt
=x/2+1/4∫(0->x)cos(2x-2t)d(2x-2t)
=x/2+1/4sin(2x-2t)|(0->x)
=x/2+1/4(sin(2x-2x)-sin(2x-2*0)
=x/2+sin2x/4
所以
d/dx∫(0->x)sin(x-t)^2dt
=d(x/2+sin2x/4)/dx
=1/2+1/4*cos2x*2
=1/2+cos2x /2
求定积分,上限为兀/4,下限为0,x/(l+cos2x)dx
∫[x/(1+cos2x)]dx
=∫[x/(1+2cos^2 x-1)]dx
=∫[x/(2cos^2 x)]dx
=(1/2)∫(x/cos^2 x)dx
=(1/2)∫x*sec^2 xdx
=(1/2)∫xd(tanx)
=(1/2)[x*tanx-∫tanxdx]
=(1/2)[x*tanx-∫(sinx/cosx)dx]
=(1/2)[x*tanx+∫(1/cosx)d(cosx)]
=(1/2)[x*tanx+ln|cosx|]
因为x∈[0,π/4],则cosx＞0
所以：原定积分=(1/2)[x*tanx+ln(cosx)]|
=(1/2){[(π/4)*1+ln(√2/2)]-[0*0+0]}
=(1/2)*[(π/4)-(1/2)ln2]
=(π/8)-(1/4)ln2
f(x)=1/(1+cos^2x)+∫(上限π,下限0)f(x)sinxdx,求f（x） 如题
令A=∫[0,π]f(x)sinxdx
注意定积分是数值
f(x)=1/(1+cos^2x)+A
两边同乘sinx得
f(x)sinx=sinx/(1+cos^2x)+Asinx
两边在[0,π]积分 得
∫[0,π]f(x)sinxdx=∫[0,π] [sinx/(1+cos^2x)+Asinx]dx
A=∫[0,π] [sinx/(1+cos^2x)+Asinx]dx
=[-arctancosx-Acosx] [0,π]
=π/2+2A
A=-π/2
f(x)=1/(1+cos^2x)-π/2
求下列函数的导数f(x)=∫上限x,下限2;(e^2x)dxf(x)=∫上限s³,下限x²;cost？
f(x)=∫上限x,下限2;(e^2x)dx
-->
f'(x)=e^2x
f(x)=∫上限s³,下限x²;costdt
-->
f'(x)=-cosx²,5,∵f(x) = ∫上限x,下限2;(e^2x)dx = 1/2*( e^2x - e^4 )
∴f'(x) = e^2x
∵f(x)=∫上限s³，下限x²;costdt = Sin s² - Sin x²
∴ f'(x) = -2x * Cos x²,2,求下列函数的导数
f(x)=∫上限x,下限2;(e^2x)dx
f(x)=∫上限s³,下限x²;costdt
