1. ## Integration Problem!

Hi guys,
I am having a hard time remembering how to find this type of integral:

integral from 0 to 3 of: (y^2)(9+(4(y^2)))^(1/2) dy

I thought it had some u-sub, but that didn't work because u would equal 9 + 4y^2 but then that would make du = 8y, which is not part of the integral. Could you please help me with this? Thanks in advance!

2. Originally Posted by collegestudent321
Hi guys,
I am having a hard time remembering how to find this type of integral:

integral from 0 to 3 of: (y^2)(9+(4(y^2)))^(1/2) dy

I thought it had some u-sub, but that didn't work because u would equal 9 + 4y^2 but then that would make du = 8y, which is not part of the integral. Could you please help me with this? Thanks in advance!
You could do a trig substitution. $\displaystyle 2y = 3 \tan t$

if you really want a u-stub, you'd probably have to do it twice. unless you have good insight...

3. ok, so 2y = 3tan t, so 2 dy = 3sec^2 t or dy = 3/2 sec^2 t, now the integral is: (9/4)tan^2 t (3(sec^2 t) dt
now im stuck again

4. * i meant (9/4)tan^2 t (3(sect)) dt
sorry

5. Originally Posted by collegestudent321
* i meant (9/4)tan^2 t (3(sect)) dt
sorry
that is not what I got.

I end up with $\displaystyle \frac {81}8 \int \tan^2 t \sec^3 t~dt$

6. oh ok, yea thats right... so then does that simplify to (81/8) integral of: sec^5 t - sec^3 t dt
or is that not the right way to do it?

7. Originally Posted by collegestudent321
oh ok, yea thats right... so then does that simplify to (81/8) integral of: sec^5 t - sec^3 t dt
or is that not the right way to do it?
that is fine. now, you begin the painful process of integrating things like $\displaystyle \sec^5 t$ and $\displaystyle \sec^3 t$! You can use integration by parts or reduction formulas.

just leave the $\displaystyle \int \sec^3 t~dt$ for a while. Apply the reduction formula to the $\displaystyle \int \sec^5 t ~dt$, and you can simplify in such a way that you only have to work on $\displaystyle \int \sec^3 t~dt$ in the end.

8. ok, so is there is an easier way to simplify that expression? or is there no way around this painful process? just curious so i can see all the different ways (if there are any)

9. Originally Posted by collegestudent321
ok, so is there is an easier way to simplify that expression? or is there no way around this painful process? just curious so i can see all the different ways (if there are any)
Use the fact (which can be derived via integration by parts) that $\displaystyle \int \sec^n x~dx = \frac 1{n - 1} \tan x \sec^{n - 2} x + \frac {n - 2}{n - 1} \int \sec^{n - 2} x ~dx$. if you use that on the first integral, you end up with

$\displaystyle \frac 14 \tan t \sec^3 t - \frac 14 \int \sec^3 t~dt$

and $\displaystyle \int \sec^3 t~dt$ is something you can do by parts, or reuse the above formula with $\displaystyle n = 3$. Or the easy way is to look up the formula for it. most texts will have the integral for $\displaystyle \sec^3 x$ listed in the book cover

10. where did the - 1/4 come from... im getting + 3/4

11. Originally Posted by collegestudent321
where did the - 1/4 come from... im getting + 3/4
...remember that $\displaystyle - \int \sec^3 t~dt$ we had before...! 3/4 - 1 = -1/4

12. oh ok, yea that makes sense

13. ok, so i got: the first integral is: (1/12)(sec t)
and the second integral is: (1/8)((sec t)(tan t) + ln(sec x + tan x))

for some reason that doesn't feel like its right, but im not sure.

14. Originally Posted by collegestudent321
ok, so i got: the first integral is: (1/12)(sec t)
and the second integral is: (1/8)((sec t)(tan t) + ln(sec x + tan x))

for some reason that doesn't feel like its right, but im not sure.
$\displaystyle \int \sec^3 t~dt = \frac 12 \sec t \tan t + \frac 12 \ln |\sec t + \tan t| + C$

just plug that in, and remember to back substitute. Your original problem was in the variable $\displaystyle x$, you must switch back to it.