Integral of root(-x^2+10x-16)dx

[jason_r]

Homework Statement

integral of root(-x^2+10x-16)dx

Homework Equations

The Attempt at a Solution

i can factor the thing inside the root to (x-8)(x-2) if i pull out the negative sign
no idea what to do after that

 

[lanedance]

could you try completing the square, then a trig substitution?

 

[lubuntu]

which root the square root? that factorization won't help you get anywhere good, but if you complete the square I think you'll find yourself with something you can work with.

 

[lanedance]

scratch the trig sub, first step should help

 

[Edwin]

Try rewriting the quadratic expression under the square root symbol in vertex form

sqrt[a]*sqrt[(x-h)^2 + k^2)] , as follows:

sqrt[-1]*sqrt[(x-5)^2 - 9], then let u = x - 5, and use the formula from a table of integrals for an integral of the form

sqrt[u^2 - a^2], for a > 0, and then make the appropriate back substitutions.


Note: the sqrt[-1] is just a constant, albeit a complex constant, so it can be pulled out side of the integral. The integral will turn out to be real-valued none-the-less.

I hope this helps.

 

[lanedance]

no need or use to pull -1 outside the squareroot

complete the square make the substitution, then integrate

 

[jason_r]

hmm I am having trouble integrating sqrt(u^2 - 9)
would i need to use integration by parts?

 

[Edwin]

That makes sense, then it would just be an integral of the form

sqrt[a^2 - u^2], for a > 0.

 

[lubuntu]

no, trig substituion

 

[Edwin]

Try the triq substitution of

u = 3*sin(v) for

sqrt[9 - u^2], here we just didn't factor out the -1.

 

[jason_r]

im lost...this is as far as i get

-(integral) sqrt(u^2 - 9)

i don't understand how i can use a trig identity to simplify that

 

[lubuntu]

where the heck did you get a cube from?

 

[jason_r]

typo lol

 

[lanedance]

Hmm, not sure how well trig sub works here

have a look at the derivative of arcsin...

 

[HallsofIvy]

Hmm, not sure how well trig sub works here

have a look at the derivative of arcsin...

The trig substitution x= 3 sin t will give the same thing as the arcsine integral.