求不定积分∫x^2·√(4-x^2)dx

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 11:17:52
求不定积分∫x^2·√(4-x^2)dx
xTn@:cye@\HH[Ԓ*H PR*V"@ jRN9ݵ@{y}POϻ+*; VX\1&k.)8b7}٣+:ަ^\&ZE,Mm f N]MA@ԉ*BKDNT%"ِ T,*@A@ reC[[J1x0?%.̹?he5Y."HP>8xxkj,e WXϚb젿N%ϴYC$v!A3)8eM1_a$.󬠝ßOV H,%GR:y|hLT~<|46|:ڄ2J9 Ϣ(~M0hfBIYJSjrKU&7P;@Ȩcd8ʓ;?n jm

求不定积分∫x^2·√(4-x^2)dx
求不定积分∫x^2·√(4-x^2)dx

求不定积分∫x^2·√(4-x^2)dx
令x=2sint,dx=2costdt
代入原式得:=∫4sint^2.2cost.2costdt
=16∫sint^2.cost^2dt
=4∫sin2t^2dt
=4∫[1-(cos4t-1)/2]dt
=6t-sin4t/2+c
t=arcsinx/2代入,得
2arcsin(x/2)-x/2*√(4-x^2)+C

第一方法:
∫x²/√(4-x²)dx (三角换元,令x=2sint)
=∫4(sint)^2/√(4(cost)^2)d(2sint)
=∫4(sint)^2/(2cost)*(2cost)dt
=∫4(sint)^2dt (倍角公式 cos2t=1-2(sint)^2)
=∫2(1-cos2t)dt
=2t-sin2...

全部展开

第一方法:
∫x²/√(4-x²)dx (三角换元,令x=2sint)
=∫4(sint)^2/√(4(cost)^2)d(2sint)
=∫4(sint)^2/(2cost)*(2cost)dt
=∫4(sint)^2dt (倍角公式 cos2t=1-2(sint)^2)
=∫2(1-cos2t)dt
=2t-sin2t+C (将 t=arcsin(x/2)带回)
=2arcsin(x/2)-2(x/2)*√(1-x^2/4)+C
=2arcsin(x/2)-x/2*√(4-x^2)+C
C为任意常数。
第二方法
>> sym x;
>> simple(int(x^2/sqrt(4-x^2),x))

simplify:

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


radsimp:

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


combine(trig):

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


factor:

-1/2*x*(-(x-2)*(x+2))^(1/2)+2*asin(1/2*x)


expand:

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


combine:

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


convert(exp):

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


convert(sincos):

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


convert(tan):

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


collect(x):

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


mwcos2sin:

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)


ans =

-1/2*x*(4-x^2)^(1/2)+2*asin(1/2*x)

ans即为答案.

收起