# Thread: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

1. ## integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

I forgot how to do these kinds of problems. Could someone please show me how to find this integral (http://www.wolframalpha.com/input/?_=1322280641770&i=integrate+sqrt%2872^2t^4+%2b+48 ^2t^2%2b16^2%29&fp=1&incTime=true) ? Wolfram Alpha doesn't show me the steps for this; it just gives a final answer.

Any help would be greatly appreciated!
Thanks in advance!

2. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

Originally Posted by s3a
I forgot how to do these kinds of problems. Could someone please show me how to find this integral (http://www.wolframalpha.com/input/?_=1322280641770&i=integrate+sqrt%2872^2t^4+%2b+48 ^2t^2%2b16^2%29&fp=1&incTime=true) ? Wolfram Alpha doesn't show me the steps for this; it just gives a final answer.

Any help would be greatly appreciated!
Thanks in advance!
I would start by completing the square...

3. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

The expression $72^2 t^4+48^2 t+16^2$ is of the form $a^2+b^2+2ab$.

$72^2 t^4+48^2 t+16^2 = (72t^2+16)^2$

So $\sqrt{(72t^2+16)^2}=?$

4. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

Prove it: I don't see how to complete the square for this. I do know how to complete the square though generally.

For example:
x^2 - 2x + y^2 = 0
(x - 2)^2 + y^2 = 0 + 1

Did you mean what sbhatnagar said? If not, could you show me your way too please?

5. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

Originally Posted by s3a
Prove it: I don't see how to complete the square for this. I do know how to complete the square though generally.

For example:
x^2 - 2x + y^2 = 0
(x - 2)^2 + y^2 = 0 + 1

Did you mean what sbhatnagar said? If not, could you show me your way too please?
If you are integrating at this level you ought to know how to complete the square on something that has the form $72^2w^2 + 48^2w + 16^2$:

complete the square 72&#94;2w&#94;2 &#43; 48&#94;2w &#43; 16&#94;2 - Wolfram|Alpha

Obviously Prove It meant what sbhatnagar said. How else would you understand "complete the square"?

6. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

Originally Posted by s3a
Prove it: I don't see how to complete the square for this. I do know how to complete the square though generally.

For example:
x^2 - 2x + y^2 = 0
(x - 1)^2 + y^2 = 0 + 1

Did you mean what sbhatnagar said? If not, could you show me your way too please?
$f(t) = 72^2t^4 + 48^2t^2 + 16^2$

Let $u = t^2$ and a quadratic will suddenly appear: $72^2u^2 + 48^2u + 16^2$. Complete the square as you would normally:

$72^2\left(u^2 + \left(\dfrac{48}{72}\right)^2u + \left(\dfrac{16}{72}\right)^2\right) = 72^2\left(u^2 + \left(\dfrac{2}{3}\right)^2u + \left(\dfrac{2}{9}\right)^2\right)$

7. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

$72^2t^4 + 48^2t^2 + 16^2 =$

$(2^6 \cdot 3^4)t^4 + (3^2 \cdot 2^8)t^2 + 2^8 =$

$2^6[3^4t^4 + (3^2 \cdot 2^2) t^2 + 2^2] =$

$2^6(3^2t^2 + 2)^2 =$

$64(9t^2+2)^2$

8. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

Hello, s3a;697799!

$\int \sqrt{ 72^2t^4+ 48^2t^2 + 16^2}\,dt$
It is unfortunate that "complete the square" was mentioned.

Factor the given polynomial:

. . $8^2\left(9^2t^4 + 6^2t^2 + 2^2\right) \:=\:64(81t^4 + 36t^2 + 4) \:=\:64(9t^2+2)^2$

The integral becomes:
. . $\int\sqrt{64(9t^2+2)^2}\,dt \;=\;\int 8(9t^2+2)\,dt \;=\;8\int(9t^2+2)\,dt$

Got it?

Edit: Too slow . . . again!
.

9. ## Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

Originally Posted by Soroban
Hello, s3a;697799!

It is unfortunate that "complete the square" was mentioned.

Factor the given polynomial:

. . $8^2\left(9^2t^4 + 6^2t^2 + 2^2\right) \:=\:64(81t^4 + 36t^2 + 4) \:=\:64(9t^2+2)^2$

The integral becomes:
. . $\int\sqrt{64(9t^2+2)^2}\,dt \;=\;\int 8(9t^2+2)\,dt \;=\;8\int(9t^2+2)\,dt$

Got it?

Edit: Too slow . . . again!
.
For anyone who wants a general method which will work for similar problems, completing the square is the best method. It is also a valid method in this case.