Question

∫［a，0］x^2·根号(a^2-x^2)dx求定积分

Answer 1

(0,a) ∫x²√(a²-x²) dx

原式=(0,a)∫(ax²√[1-(x/a)²]dx

令x/a=sint,则dx=acostdt,x=0时,t=0；x=a时,t=π/2.

故原式=(0,π/2)a⁴∫sin²tcos²tdt=(0,π/2)(a⁴/4)∫sin²(2t)dt=(0,π/2)(a⁴/8)∫sin²(2t)d(2t)

=(0,π/2)(a⁴/16)∫[(1-cos4t)/2]d(4t)=(0,π/2)(a⁴/32)∫[(1-cos4t)d(4t)

=[(a⁴/32)(4t-sin4t)](0,π/2)=(a⁴/32)×(2π)=πa⁴/16

扩展资料

常用积分公式：

1）∫0dx=c

2）∫x^udx=(x^(u+1))/(u+1)+c

3）∫1/xdx=ln|x|+c

4）∫a^xdx=(a^x)/lna+c

5）∫e^xdx=e^x+c

6）∫sinxdx=-cosx+c

7）∫cosxdx=sinx+c

Answer 2

令x=√2sina则dx=√2cosada√(2-x²)=cosax=0,a=0x=√2,a=π/2所以原式=∫(0-π/2）2sin²a*cosa*√2cosada=∫(0-π/2）2√2sin²acos²ada=√2/2*∫(0-π/2）sin²2ada=√2/4*∫(0-π/2）(1-cos2a)/2d2a=√2/4(2a-sin2a)/2(0-π/2）=√2/4*(π-0)=π√2/4

Answer 3

详情如图所示

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