Question

What is the integral `int_0^(2sqrt(3)) x^3/sqrt(16 - x^2) dx`?

Answer 1

`40/3`

Explanation:

Use the trig substitution `x = 4sintheta`. Then `dx = 4costhetad theta`. We adjust the bounds of integration accordingly.

`int_0^sin(2sqrt(3)) (4sintheta)^3/sqrt(16 - (4sintheta)^2) * 4costheta d theta`

`int_0^sin(2sqrt(3)) (64sin^3theta)/sqrt(16 - 16sin^2theta) * 4costheta d theta`

`int_0^sin(2sqrt(3)) (64sin^3theta)/sqrt(16(1 - sin^2theta)) * 4costheta d theta`

`int_0^sin(2sqrt(3)) (64sin^3theta)/sqrt(16cos^2theta) * 4costheta d theta`

`int_0^sin(2sqrt(3)) (64sin^3theta)/(4costheta) * 4costheta d theta`

`int_0^sin(2sqrt(3)) 64sin^3theta d theta`

`int_0^sin(2sqrt(3)) 64(1 - cos^2theta)sin theta d theta`

Now let `u = costheta`. Then `du = -sintheta d theta` and `d theta = (du)/(-sintheta)`.

`int_1^cos(sin(2sqrt(3))) 64(1 - u^2)sintheta * (du)/(-sintheta)`

`-int_1^cos(sin(2sqrt(3))) 64(1 - u^2)du`

`-int_1^cos(sin(2sqrt(3))) 64 - 64u^2 du`

`-[64u - 64/3u^3]_1^cos(sin(2sqrt(3))`

`-[64costheta - 64/3cos^3theta]_0^sin(2sqrt(3))`

`-[16sqrt(16 - x^2) - 1/3(16 - x^2)^(3/2)]_0^(2sqrt(3)`

`-(16sqrt(16 - (2sqrt(3))^2) - 1/3sqrt(16 - (2sqrt(3))^2)^(3/2) - (16sqrt(16 - 0) - 1/3(16 - 0)^(3/2)))`

`-16sqrt(4) + 1/3(4)^(3/2) + 16(4) - 1/3(64)`

`-32 + 8/3 + 64 - 64/3`

`40/3`

Hopefully this helps!