请问定积分(0,a)∫x^2√(a^2-x^2) dx怎么做？
\u5b9a\u79ef\u5206(0\u5230a) \u222bx^2*(\u221a[(a - x)/(a + x)] dx\u4ee4x = asinz\uff0cdx = acosz
x = 0 => z = 0
x = a => sinz = 1 => z = \u03c0/2
\u222b(0\u2192a) x²\u221a[(a - x)/(a + x)] dx
= \u222b(0\u2192a) x² * [\u221a(a - x)\u221a(a - x)]/[\u221a(a + x)\u221a(a - x)] dx
= \u222b(0\u2192a) x² * (a - x)/\u221a(a² - x²) dx
= \u222b(0\u2192\u03c0/2) (asinz)²(a - asinz)/(acosz) * (acosz dz)
= \u222b(0\u2192\u03c0/2) (a³sin²z)(1 - sinz) dz
= a³\u222b(0\u2192\u03c0/2) sin²z dz - a³\u222b(0\u2192\u03c0/2) sin³z dz
= (a³/2)\u222b(0\u2192\u03c0/2) (1 - cos2z) dz - a³\u222b(0\u2192\u03c0/2) (cos²z - 1) d(cosz)
= (a³/2)[z - (1/2)sin2z] |(0\u2192\u03c0/2) - a³[(1/3)cos³z - cosz] |(0\u2192\u03c0/2)
= (a³/2)(\u03c0/2) - a³[((1/3)(- 1) - (- 1)) - ((1/3)(1) - 1)]
= a³(\u03c0 - 4a)/4
\u7a0d\u7b49\uff0c\u56fe\u7247\u5df2\u7ecf\u4f20\u4e0a\u3002
\u70b9\u51fb\u653e\u5927\uff0c\u518d\u70b9\u51fb\u518d\u653e\u5927\u3002
令x=asint
则dx=acost dt
代入原式，注意积分上下限变为0到PI
查表积分可以得出
结果为（pi*a^2)/4
令x
=
asinθ，dx
=
acosθdθ
原式=
∫(0→π/2)
(acosθ)/(asinθ
+
acosθ)
dθ
=
(1/2)∫(0→π/2)
2cosθ/(sinθ
+
cosθ)
dθ
=
(1/2)∫(0→π/2)
[(sinθ
+
cosθ)
-
(sinθ
-
cosθ)]/(sinθ
+
cosθ)
dθ
=
(1/2)∫(0→π/2)
dθ
-
(1/2)∫(0→π/2)
(sinθ
-
cosθ)/(sinθ
+
cosθ)
dθ
=
(1/2)(π/2)
-
(1/2)∫(0→π/2)
-
d(cosθ
+
sinθ)/(sinθ
+
cosθ)
dθ
=
π/4
+
(1/2)ln(sinθ
+
cosθ)
|(0→π/2)
=
π/4
+
(1/2)[ln(1
+
0)
-
ln(0
+
1)]
=
π/4
难啊。
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