∫arcsinxdx分部积分
@申询3604：∫arctanxdx用分部积分法求解 - 作业帮
倪奇18695736704…… [答案] ∫ arctanx dx = x * arctanx - ∫ x d(arctanx) = x * arctanx - ∫ x/(1+x²) dx = x * arctanx - (1/2)∫ d(x²)/(1+x²) = x * arctanx - (1/2)∫ d(1+x²)/(1+x²) = x * arctanx - (1/2)ln(1+x²) + C
@申询3604：计算不定积分 ∫arcsin xdx - 作业帮
倪奇18695736704…… [答案] ∫arcsin xdx(分部积分法) =xarcsinx-积分:xd(arcsinx) =xarcsinx-积分:x/根号(1-x^2)dx =xarcsinx+1/2积分:d(1-x^2)/根号(1-x^2) =xarcsinx+1/2*2根号(1-x^2)+C =xarcsinx+根号(1-x^2)+C (C为常数)
@申询3604：求解答,微积分∫arcsinxdx要详细步骤 -
倪奇18695736704…… ∫arcsinxdx= xarcsinx + √(1-x²) +C.C为常数. 用分部积分法:∫ u dv = uv - ∫ v du ∫ arcsinx dx = x arcsinx - ∫ x darcsinx = xarcsinx - ∫ x / √(1 - x²) dx = xarcsinx + 1/2 ∫ 1/√(1-x²) d(1-x²) = xarcsinx + √(1-x²) +C 扩展资料: 求不定积分的方法...
@申询3604：求∫arcsin√xdx -
倪奇18695736704…… 令arcsin√x=t √x=sint x=sin^2 t dx=dsin^2 t 原式=∫tdsin^2 t =tsin^2 t-∫sin^2 t dt =tsin^2 t-∫(1-cos2t)dt/2 = tsin^2 t-1/2∫dt+1/2∫cos2tdt = tsin^2 t-t/2+1/4∫cos2td2t = tsin^2 t-t/2+sin2t/4+C =xarcsin√x-arcsin√x/2+2sintcost/4+C =xarcsin√x-arcsin...
@申询3604：分部积分法求 ∫xarccosxdx - 作业帮
倪奇18695736704…… [答案] ∫ x · arccos(x) dx = ∫ arccos(x) d(x²/2) = (1/2)x² · arccos(x) - (1/2)∫ x² d(arccos(x)) = (1/2)x² · arccos(x) - (1/2)∫ x² ·... arccos(x) + (1/2)∫ (1 - cos2θ)/2 dθ = (1/2)x² · arccos(x) + (1/4)(θ - 1/2 · sin2θ) + C = (1/2)x² · arccos(x) + (1/4)arcsin(x) - ...
@申询3604：求不定积分∫(arcsinx)2dx. - 作业帮
倪奇18695736704…… [答案] ∫(arcsinx)2dx=x(arcsinx)2-∫xd(arcsinx)2 =x(arcsinx)2+∫ 2xarcsinx 1−x2dx =x(arcsinx)2+2∫arcsinxd 1−x2 =x(arcsinx)2+2 1−x2arcsinx−2∫dx =x(arcsinx)2+2 1−x2arcsinx−2x+C,其中C为任意常数.
@申询3604：用分部积分法做∫(arcsin√x/√x)dx 我知道真确答案是2√xarcsin√x+2√(1 - x)+c - 作业帮
倪奇18695736704…… [答案] ∫(arcsin√x/√x)dx 因 2d√x=dx /√x = 2∫arcsin√xd√x 令√x=u = 2∫arcsin u du =2 {u arcsin u- ∫u/√[1-u^2]du} =2 u arcsin u+ ∫1/√[1-u^2]d(1-u^2) =2 u arcsin u+ 2√[1-u^2]+c =2 √x arcsin √x+ 2√[1-x]+c
@申询3604：∫xcos2xdx用分部积分怎么解 - 作业帮
倪奇18695736704…… [答案] ∫ xcos2x dx = (1/2)∫ xcos2x d(2x) = (1/2)∫ x d(sin2x) = (x/2)sin2x - (1/2)∫ sin2x dx = (x/2)sin2x - (1/2)(1/2)∫ sin2x d(2x) = (x/2)sin2x + (1/4)cos2x + C
@申询3604：用分部积分法求∫(arcsinx)2dx,要步骤
倪奇18695736704…… ∫(arcsinx)2dx = x(arcsinx)² - ∫xd(arcsinx)²= x(arcsinx)² - ∫2xarcsinx*1/√(1-x²)dx= x(arcsinx)² ∫arcsinx*1/√(1-x²)d(1-x²)= x(arcsinx)² 2∫arcsinx*1/2√(1-x²)d(1-x²)...
@申询3604：用分部积分法求:∫xarcsinxdx -
倪奇18695736704…… 解:∫xarcsinxdx=1/2*∫arcsinxdx^2=1/2*x^2*arcsinx-1/2∫x^2darcsinx=1/2*x^2*arcsinx-1/2∫x^2/√(1-x^2)dx 令x=sint,那么,∫x^2/√(1-x^2)dx=∫(sint)^2/costdsint=∫(sint)^2dt=∫(1-cos2t)/2dt=1/2t-1/4sin2t+C=1/2t-1/2sint*cost+C 又x=sint,则t=arcsinx,...
倪奇18695736704…… [答案] ∫ arctanx dx = x * arctanx - ∫ x d(arctanx) = x * arctanx - ∫ x/(1+x²) dx = x * arctanx - (1/2)∫ d(x²)/(1+x²) = x * arctanx - (1/2)∫ d(1+x²)/(1+x²) = x * arctanx - (1/2)ln(1+x²) + C
@申询3604：计算不定积分 ∫arcsin xdx - 作业帮
倪奇18695736704…… [答案] ∫arcsin xdx(分部积分法) =xarcsinx-积分:xd(arcsinx) =xarcsinx-积分:x/根号(1-x^2)dx =xarcsinx+1/2积分:d(1-x^2)/根号(1-x^2) =xarcsinx+1/2*2根号(1-x^2)+C =xarcsinx+根号(1-x^2)+C (C为常数)
@申询3604：求解答,微积分∫arcsinxdx要详细步骤 -
倪奇18695736704…… ∫arcsinxdx= xarcsinx + √(1-x²) +C.C为常数. 用分部积分法:∫ u dv = uv - ∫ v du ∫ arcsinx dx = x arcsinx - ∫ x darcsinx = xarcsinx - ∫ x / √(1 - x²) dx = xarcsinx + 1/2 ∫ 1/√(1-x²) d(1-x²) = xarcsinx + √(1-x²) +C 扩展资料: 求不定积分的方法...
@申询3604：求∫arcsin√xdx -
倪奇18695736704…… 令arcsin√x=t √x=sint x=sin^2 t dx=dsin^2 t 原式=∫tdsin^2 t =tsin^2 t-∫sin^2 t dt =tsin^2 t-∫(1-cos2t)dt/2 = tsin^2 t-1/2∫dt+1/2∫cos2tdt = tsin^2 t-t/2+1/4∫cos2td2t = tsin^2 t-t/2+sin2t/4+C =xarcsin√x-arcsin√x/2+2sintcost/4+C =xarcsin√x-arcsin...
@申询3604：分部积分法求 ∫xarccosxdx - 作业帮
倪奇18695736704…… [答案] ∫ x · arccos(x) dx = ∫ arccos(x) d(x²/2) = (1/2)x² · arccos(x) - (1/2)∫ x² d(arccos(x)) = (1/2)x² · arccos(x) - (1/2)∫ x² ·... arccos(x) + (1/2)∫ (1 - cos2θ)/2 dθ = (1/2)x² · arccos(x) + (1/4)(θ - 1/2 · sin2θ) + C = (1/2)x² · arccos(x) + (1/4)arcsin(x) - ...
@申询3604：求不定积分∫(arcsinx)2dx. - 作业帮
倪奇18695736704…… [答案] ∫(arcsinx)2dx=x(arcsinx)2-∫xd(arcsinx)2 =x(arcsinx)2+∫ 2xarcsinx 1−x2dx =x(arcsinx)2+2∫arcsinxd 1−x2 =x(arcsinx)2+2 1−x2arcsinx−2∫dx =x(arcsinx)2+2 1−x2arcsinx−2x+C,其中C为任意常数.
@申询3604：用分部积分法做∫(arcsin√x/√x)dx 我知道真确答案是2√xarcsin√x+2√(1 - x)+c - 作业帮
倪奇18695736704…… [答案] ∫(arcsin√x/√x)dx 因 2d√x=dx /√x = 2∫arcsin√xd√x 令√x=u = 2∫arcsin u du =2 {u arcsin u- ∫u/√[1-u^2]du} =2 u arcsin u+ ∫1/√[1-u^2]d(1-u^2) =2 u arcsin u+ 2√[1-u^2]+c =2 √x arcsin √x+ 2√[1-x]+c
@申询3604：∫xcos2xdx用分部积分怎么解 - 作业帮
倪奇18695736704…… [答案] ∫ xcos2x dx = (1/2)∫ xcos2x d(2x) = (1/2)∫ x d(sin2x) = (x/2)sin2x - (1/2)∫ sin2x dx = (x/2)sin2x - (1/2)(1/2)∫ sin2x d(2x) = (x/2)sin2x + (1/4)cos2x + C
@申询3604：用分部积分法求∫(arcsinx)2dx,要步骤
倪奇18695736704…… ∫(arcsinx)2dx = x(arcsinx)² - ∫xd(arcsinx)²= x(arcsinx)² - ∫2xarcsinx*1/√(1-x²)dx= x(arcsinx)² ∫arcsinx*1/√(1-x²)d(1-x²)= x(arcsinx)² 2∫arcsinx*1/2√(1-x²)d(1-x²)...
@申询3604：用分部积分法求:∫xarcsinxdx -
倪奇18695736704…… 解:∫xarcsinxdx=1/2*∫arcsinxdx^2=1/2*x^2*arcsinx-1/2∫x^2darcsinx=1/2*x^2*arcsinx-1/2∫x^2/√(1-x^2)dx 令x=sint,那么,∫x^2/√(1-x^2)dx=∫(sint)^2/costdsint=∫(sint)^2dt=∫(1-cos2t)/2dt=1/2t-1/4sin2t+C=1/2t-1/2sint*cost+C 又x=sint,则t=arcsinx,...
