How can the integral of the square root of t squared plus 9 be evaluated?

[Aerosion]

Homework Statement

\int \sqrt{t^2+9}

Homework Equations

The Attempt at a Solution

Apparently you can't solve this equation as you would \int \sqrt{t+9}, which would come out to \frac{2(t+9)^3/2 }{3}.

Instead, my calculator is getting this extremely complicated answer involving ln functions, and therefore I don't know how to integrate this.

I know I'm not giving much to go on as far as an attempt on a solution is concerned, but I really don't have a definite way to begin, and I need to know how to evaluate this.

 

[gammamcc]

Trig sub...

Think of a right triangle with one side length sqrt(9) and the other t. Pick correct sides so the radical in the integral represent something that makes sense (hypotenuse in this case). Oh ya, what would dt be?

 

[Gib Z]

Its a fatal error when you don't include your differential.

\int \sqrt{t^2+9} dt!

By trig sub, let t=3tan x

Only problem is, then you have 9\int \sec^3 x dx Left to integrate, which I can't do.

 

[Mandanesss]

To integrate the integral of (sec(x))^3 dx you'd need to factor out a sec(x) and use integration by parts letting u = sec (x) and dv = (sec(x))^2 dx. Using some simple trig manipulation you should be able to evaluate the integral. If not, let me know.

 

[Hootenanny]

Perhaps a more straight forward substutition would be let t=sinh(x), hardly any manipulation required and no integration by parts.

 

[dextercioby]

Make that t=3\sinh x plus using the formula

\cosh^2 x =\frac{\cosh 2x +1}{2}

 

[Hootenanny]

Make that t=3\sinh x plus using the formula

\cosh^2 t =\frac{\cosh 2t +1}{2}

Oops, forgot the factor of three, good catch dexter

 

[abouchard]

I hate to dig this back up, but I'm trying to do something similar to this, and I'm not having much luck. I don't know if it's because I'm not that familiar with hyperbolic trig functions or just rusty on integration.

Using the trig substitution suggested with the equation provided, I get this:
\int \sqrt{ (3 \sinh ( x ) ) ^2+9 } (3 \cosh (x) ) dx

From there, I can get it down to
9 \int \sqrt{ \cosh ^2 (x) \sinh ^2 (x) + \cosh ^2 (x)} dx

From there, I'm at a bit of a loss. I've been looking at some hyperbolic trig identities, but haven't found anything that really looks helpful. Did I multiply something out incorrectly or something stupid like that?

Sorry if the equations above don't work - I've been trying to get them to register as LaTeX, but it's not working.

 

[abouchard]

Okay, scratch that. After finding substitutions for both \sinh ^2 and \cosh ^2, I got this:

\frac{9}{2} \int \cosh (2x) + 1 dx

If I remember my rules correctly (and I'm not doing anything stupid), this is

\frac{9}{4} \sinh (2x) + x

Is there a better way to simplify this than

\frac{9}{4} 2 \sinh (\sinh ^{-1}(t/3)) \cosh (\sinh ^{-1}(t/3)) + \sinh ^{-1}(t/3)