Solving Trig Sub in Homework: Integral of (16x-2x^2-23)^1/2 dx

[moloko]

Homework Statement

integral of (16x-2x^2-23)^1/2 dx

Homework Equations

The Attempt at a Solution

First off it seems clear that I must complete the square within the radical.
This gives ((x-4)^2-4.5)^1/2

From there I seem to be lost, fumbling through my book for any kind of help. Can I use a trig integration, such as using the form (a^2-x^2)^1/2 to substitute x=asecT? I don't think so because (x-4)^2 cannot be treated like x^2.

I also cannot find any possible u-sub that will work because of the radical over the entire problem.

Any type of guidance is very much appreciated!

 

[siddharth]

I don't think so because (x-4)^2 cannot be treated like x^2.

Set x-4 as t, and integrate wrt t.

 

[moloko]

Some work...

integral signs before everything

a=sqrt(4.5)
x-4 = asecT
dx=asecTtanTdTsqrt[-2*(asecT)^2-4.5] * sqrt[4.5] * secTtanTdT

sqrt[-2*(4.5sec^2(T)-4.5)*4.5] * secTtanTdT

sqrt[(-40.5)(tan^2(T)] * secTtanTdT

Subbing T=arcsec((x-4)/a)

sqrt[(-40.5)(tan^2(arcsec(T))] * sec(arcsec(T))tan(arcsec(T))dT

..heres where I start to get lost
I know sec(arcsec(t)) is just T.
I know you can do a case for tan(arcsec(T))..
if x>=1 =) sqrt(x^2-1)
or x<1 =) -sqrt(x^2-1)

though I don't think that helps much.

I'm very lost and desperate for some help, any advice is appreciated. Thank you.

 

[Dick]

Try putting (x-4)=a*sin(T). Since sin^2(T)-1=-cos^2(T) this will take care of the awkward negative sign under the integral and make your life overall much easier.