∫sinx+2dx
别雷元4653求∫1/(2+sinx)dx的不定积分 -
沈虾欣18997227073 ______ ∫1/(2+sinx)dx=2√3/3*arctan{[2√3tan(x/2)+√3]/3}+C.C为常数. 2+sinx=2sin(x/2)^2+2cos(x/2)^2+2sin(x/2)cos(x/2) dx/(2+sinx)=sec(x/2)^2dx/[2+2tan(x/2)^2+2tan(x/2)] =d(tan(x/2))/[1+tan(x/2)+tan(x/2)^2] 令u=tan(x/2) 原积分=∫du/(1+u+u^2) =∫d(u+...
别雷元4653∫1/(2+sinx)dx -
沈虾欣18997227073 ______ 2+sinx=2sin(x/2)^2+2cos(x/2)^2+2sin(x/2)cos(x/2) dx/(2+sinx)=sec(x/2)^2dx/[2+2tan(x/2)^2+2tan(x/2)]=d(tan(x/2))/[1+tan(x/2)+tan(x/2)^2] 令u=tan(x/2) 原积分=∫du/(1+u+u^2)=∫d(u+1/2)/[3/4+(u+1/2)^2](用∫dx/(a^2+x^2)公式,取a=√3/2)=1/a*...
别雷元4653∫sinx/1+x^2+(cosx)^2dx -
沈虾欣18997227073 ______ ∫(sinx+x^2+(cosx)^2)dx=∫sinxdx+∫x^2dx+∫(cosx)^2dx=-cosx+x³/3+∫(1+cos2x)d2x=-cosx+x³/3+2x+sin2x
别雷元4653∫sinx/1 sinxdx怎样求 -
沈虾欣18997227073 ______ ∫[sinx/(1+ sinx) ]dx =∫[1 - 1/(1+ sinx) ]dx = x- ∫dx/(1+ sinx) = x- ∫(1-sinx)/(cosx)^2 dx = x- ∫(secx)^2 dx +∫ sinx/(cosx)^2 dx =x - tanx + (1/cosx) + C
别雷元4653求定积分∫根号1+cos2xdx,积分上限是π,积分下限是0的值? -
沈虾欣18997227073 ______ ∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√ ∵cos2x=2cos²x-1 ∴∫√(1+cos2x)dx=∫√2|cosx|dx ∴(0,π)∫√(1+cos2x)dx =(0,π/2)∫√2cosxdx+(π/2,π)∫-√2cosxdx =2√2 所以∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√2. ...
别雷元4653求∫dx/(2+cosx)sinx采纳有加分 -
沈虾欣18997227073 ______ ∫dx/(2+cosx)sinx=∫dx/(1+1+cosx)2sinx/2cosx/2 =∫sinx/2dx/(1+2(cosx/2)^2)2(sinx/2)^2cosx/2 = -∫dcosx/2/(1+2(cosx/2)^2)(1-(cosx/2)^2)cosx/2 = -∫du/(1+2u^2)(1-u^2)u (令u=cosx/2,化成了有理函数的积分)
别雷元46531+sinx/(1 - sinx)不定积分 -
沈虾欣18997227073 ______ ∫(1+sinx)/(1-sinx)=-x+2tanx+2/cosx+C.C为积分常数. 解答过程如下: (1+sinx)/(1-sinx)=-(1-sinx-2)/(1-sinx) =-1+2/(1-sinx) =-1+2(1+sinx)/cos^2x =-1+2sec^2x+2sinx/cos^2x 原式=∫(-1+2sec^2x+2sinx/cos^2x)dx =-x+2tanx-∫2/cos^2x)...
别雷元4653∫1, - 1(sinx+2)根号1 - x^2dx 急急急!!! -
沈虾欣18997227073 ______ 答:(-1→1) ∫ (sinx+2)√(1-x^2) dx=(-1→1) ∫ sinx√(1-x^2)dx+ (-1→1) ∫ 2√(1-x^2) dx=0+(-1→1) 2∫ √(1-x^2) dx,设x=sint=(-π/2→π/2) 2∫ cost d(sint)=(-π/2→π/2) ∫ (cos2t+1) dt=(-π/2→π/2) [(1/2)sin2t+t]=(0+π/2)-(0-π/2)=π
别雷元4653求不定积分∫sinxdx/(sinx+1) -
沈虾欣18997227073 ______[答案] ∫sinxdx/(sinx+1)=∫(1-1/(1+sinx)dx=∫(1-(1-sinx)/cos^2x)dx =x-∫sec^2xdx-∫1/cos^2xdcosx =x-tanx+1/cosx+C(常数)