Tính: `\int_{-2}^{2} (x^3 cos` `x/2+1/2)\sqrt(4-x^2)` `dx`

`I = int_( - 2)^2(x ^3 cos \fracx2 + 1/2)sqrt(4 - x ^2) dx`

`= int_( - 2)^2 \underbrace{x^3 cos fracx2sqrt(4 - x ^2)}_{f(x)} dx + underbrace(1/2int_( - 2)^2sqrt(4 - x ^2) dx)_(I_2) `

Ta có: `f(-x)= (-x)^3 cos fracx 2sqrt(4 - (-x) ^2)`

`=-(x^3 cos frac x 2sqrt(4 - x ^2))=-f(x)`

`=>f(x)` là hàm số lẻ

`=>int_( - 2)^2f(x)dx=0`

Với: `I _2 = 1/2int_( - 2)^2sqrt(4 - x ^2) dx`

Đặt `x=2sint=>dx=2costdt`

Đổi cận: `{(x=-2=>t=-pi/2),(x=2=>t=pi/2):}`

Khi đó, `I=int_(-pi/2)^(pi/2)costsqrt(4-4sin^2t)dt`

`=2int_(-pi/2)^(pi/2)costsqrt(cos^2t)dt`

`=2int_(-pi/2)^(pi/2)cos^2tdt`

`=2int_(-pi/2)^(pi/2)((cos2t)/2+1/2)dt`

`2[(sin2t)/4+t/2]_(-pi/2)^(pi/2)`

`=pi`

Do đó, `I = int_(-2)^2f(x)dx+I_2=0+pi=pi`

Vậy `I=pi`