Is Trigonometric Substitution Needed for ∫ √9-2(x-1)²?

[cogs24]

hi guys
im debating whether this question requires trignometric substitution or just normal substitution.

∫ √9-2(x-1)²

Im leaning towards normal substitution, with u = x-1, but I am not sure
Any ideas
Thanx heaps

 

[whozum]

Is it

\int \sqrt{9-(2x-1)^2}?

If so a normal one won't do, youll need a trig one.

 

[dextercioby]

Try to bring it to the form

C\int\sqrt{1-(something)^2} \ dx

,where C is a constant.

Daniel.

 

[cogs24]

2 is on the outside of the (x+1)

 

[cogs24]

(x-1) sorry

 

[Jameson]

Well if you substitute u = x-1, you get the integral...

\int \sqrt{9 - 2u^2}du

I think trig substitution is where this is leading.

 

[GCT]

yeah from what I remember, trig substitution was the way to go.

\sqrt(2) \int \sqrt{(9/2-u^{2})} ~du

 

[cogs24]

hi guys
im still stuck on this question, it has really got me stumped
i know that for the form

√a² - x².dx

x=asinθ, dx = acosθ.dθ

but i just can't seem to put it together

 

[dextercioby]

Why not.U've got an |a| coming out of the sqrt and u'll get a square

\int \sqrt{a^{2}-x^{2}} \ dx =a|a|\int \cos^{2}\theta \ d\theta

Then use the double angle formula.

Daniel.

 

[cogs24]

the 2 outside the (x-1)^2 is confusing me
Im not sure what my x and a squared is?

 

[dextercioby]

Look at post #6.If you make the substitution

\sqrt{2}u=t,

u can put that integral under the form

\frac{1}{\sqrt{2}} \int \sqrt{3^{2}-t^{2}} \ dt

And now do that substitution involving "sin".

Daniel.

 

[cogs24]

so, sqrt 2u = sin theta

 

[dextercioby]

Nope.u\sqrt{2}=3\sin\theta

Daniel.