∫x^3 sin^2xdx
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/27 19:23:34
xuRNP[
aQظHPt%>bHLpenXȂ7($<¹/
Uw3gΜ9zWo:G̀/j@EP0`ؐA!M(m\,cHIY'vʰP/>Xf
+F6>EwF=YKBCf*LPh6
)y*9-"8Hq֔BZEpv9v^|wOwe~]?o7Uw>`,#oP }~W:SCg2utQ.u~AӅ&by)'w0B4C- ]C;ªmC8ɆyVUm!
∫x^3 sin^2xdx
∫x^3 sin^2xdx
∫x^3 sin^2xdx
∫ x^3sin^2x dx
= ∫ x^3 * (1 - cos(2x))/2 dx
= (1/8)x^4 - (1/2)K
where K = ∫ x^3cos(2x) dx
令f = x^3 and g = cos(2x),以下分别对f求导,对g求积分
f = x^3 g = cos2x
f' = 3x^2 g(1) = (1/2)sin(2x)
f'' = 6x g(2) = (- 1/4)cos(2x)
f''' = 6 g(3) = (- 1/8)sin(2x)
f'''' = 0 g(4) = (1/16)cos(2x)、这个是分部积分法的速解法
于是∫ x^3cos(2x) dx
= [f * g(1)] - [f' * g(2)] + [f'' * g(3)] - [f''' * g(4)]
= [x^3 * (1/2)sin(2x)] - [3x^2 * (- 1/4)cos(2x)] + [6x * (- 1/8)sin(2x)] - [6 * (1/16)cos(2x)]
= (1/2)x^3sin(2x) + (3/4)x^2cos(2x) - (3/4)xsin(2x) - (3/8)cos(2x)
原式 = (1/8)x^4 - (1/4)x^3sin(2x) - (3/8)x^2cos(2x) + (3/8)xsin(2x) + (3/16)cos(2x) + C
∫x^3 sin^2xdx
∫x/sin^2xdx
求不定积分 ∫(e^2x)sin 3xdx
求不定积分 ∫(e^2x)sin 3xdx
求不定积分 ∫x sin 3xdx
∫sin^2x/xdx求其不定积分
∫sin^2√x/√xdx
∫sin^2xdx
求不定积分 ∫xdx/sin^2x 求不定积分 ∫xdx/(sin^2)x
(xdx)/sin^2x
∫sin x /sin 2xdx 求详解,
∫sin^3xcos^2xdx
求∫cosx/sin^2xdx; ∫sec5xdx; ∫1/(√x+3√x)dx; ∫[√﹙x-1﹚/x]dx; ∫sin√xdx;
∫sin^3 xcos xdx.
∫sin^2 xdx=?
∫sin^2xdx/(1+cos^2x)求积分
∫sin^3x/cos^4xdx 求积分,
∫sin^4xdx/cos^2x 求积分.坐等