Question

求不定积分∫x^2·√(4-x^2)dx

Answer 1

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.

Answer 2

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