Question

Using trigonometric substitution, how do you integrate integral of `x^3 (x^2+4)^(1/2)`?

Answer 1

`intx^3(x^2+4)^(1/2)dx=1/5(4+x^2)^2sqrt(4+x^2)-4/3(4+x^2)sqrt(4+x^2)`

See explanations below.

Explanation:

`intx^3(x^2+4)^(1/2)dx=intx^3sqrt((x^2+4))dx`

let `x=2tan(theta)`

`dx=2sec(theta)^2d theta`

So :

`intx^3(x^2+4)^(1/2)dx=int16tan(theta)^3sqrt((4tan(theta)^2+4))sec(theta)^2d theta`

`=16inttan(theta)^3sqrt(4(tan(theta)^2+1))sec(theta)^2d theta`

`=32inttan(theta)^3cancel(sqrt(tan(theta)^2+1))^(color(red)(=sec(theta)))sec(theta)^2d theta`

`=32inttan(theta)^3sec(theta)^3d theta`

`=32int(sin(theta)^3)/(cos(theta)^6)d theta`

`=-32int(-sin(theta)sin(theta)^2)/(cos(theta)^6)d theta`

`=-32int(-sin(theta)(1-cos(theta)^2))/(cos(theta)^6)d theta`

let `u=cos(theta)`
`du=-sin(theta)d theta`

So : `32inttan(theta)^3sec(theta)^3d theta=-32int(1-u^2)/(u^6)du`

`=32intu^(-4)du-32intu^(-6)du`

`=32/5u^(-5)-32/3u^(-3)`

`=32/5cos(theta)^(-5)-32/3cos(theta)^-3`

Finally, because `theta=arctan(x/2)` and `cos(arctan(x/2))=2/sqrt(4+x^2)`,

`intx^3(x^2+4)^(1/2)dx=1/5(4+x^2)^2sqrt(4+x^2)-4/3(4+x^2)sqrt(4+x^2)`

\0/ Here's our answer !