Question

请问定积分(0,a)∫x^2√(a^2-x^2) dx怎么做？

请问定积分(0,a)∫x^2√(a^2-x^2) dx怎么做？

Answer 1

三角换元。
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

Answer 2

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

Answer 3

难啊。