Trigonometric Substitution for Integration?

Question

The problem is the integral of x ( 3x^2 - 18x + 20 )^(1/2)

I tried completing the square and got the integral of ((x-3)^2-7/3)^(1/2)

I then set x = (7/3)^(1/2)*sec(theta)
I don't think I did this part right and need to know what I should substitute for x so I can integrate.

The answer can't be 3, it has to be in the form of another variable, theta , and it has to be in the form of a trig fuction so I can get rid of the square root. I think it should be secant function?

You can't complete the square if the coefficent is 3, I factored it out and set it outside the integral as the square root of 3.

This is what I did and I'm not sure what I did wrong.. probably algebra error.

Integral ( x ( 3x^2 - 18x + 20)^(1/2)

Integral (x * 3^(1/2) ( x^2 - 6x + 20/3)^(1/2)

3^(1/2) Integral (x ( x^2 - 6x + 9 + 20/3 - 9) ^ (1/2)

I'm just going to completely ignore everything outside the integral and the integral itself and focus on what is in the square root.

x^2 - 6x + 9 + 20/3 - 9
(x-3)^2 - 7/3

x-3 = (7/3)^(1/2) * sec (theta)
x = (7/3)^(1/2) * sec (theta) + 3

Omg I know what I did wrong. I put - 3 instead of +3 before I tried to integrate. -_- thanks

Answer 1

3x^2 - 18x + 20 = 3 (x-3)^2 - 7

if x = (7/3)^(1/2)*sec(theta)
then dx = (7/3)^.5 * sec(theta) tan (theta) d-theta

your integrand needs to be 1-sec^2 (theta) to get tan^2(theta)

your sub does not work.

you need to set x - 3 = (7/3)^(1/2)*sec(theta)
so that you can do sqrt of perfect square

Answer 2

((x-3)^2-7/3)^(1/2) this can be easily integrated, put y = x-3.

Answer 3

It can be done, but involves a lot of work. You can email me & I'll give you the solution tomorrow