# Integration Help

∫(9+16x2)3/2dx

## The Attempt at a Solution

I have tried to solve this one in various ways. At first, I tried using trig substitutions and then integration by parts. I attached a pdf of what I tried to do. I'm unsure as to what I am doing wrong. Is there something I am missing here? I tried to do a u substitution with u=√9+16x2 but that seemed to not work either. Other ways just get me bogged down in a very long integration by parts thing. Is there any other way? Can anyone shed some light on what I can do to solve this? Thanks!

#### Attachments

• Math.pdf
1.7 MB · Views: 182

jfizzix
Gold Member
It seems like trig substitutions are the way to go. I looked at it quickly, and if you can integrate $sec^{5}(\theta)$, you can solve the problem

That's what I kept getting to. However, it seemed like it was going to take many integration by parts. I guess there is no other option. Any advice on how to integrate sec^5 (x). I saw somewhere else to write it as
cos (x) / cos^6 (x)

verty
Homework Helper
You can do it by parts, it is ugly though.

Let d(sec) = sec(x) tan(x) dx
So int sec tan^2 dx = int tan d(sec)

I'll use this to make the structure easier to see.

sec^5 = sec (1 + tan^2)^2
= sec + 2 sec tan^2 + sec tan^4

int sec^5 dx = int sec dx + 2 int tan d(sec) + int tan^3 d(sec)

int tan^3 d(sec) = sec tan^3 - int sec 3 tan^2 sec^2 dx
= sec tan^3 - 3 int sec^3 tan^2 dx
= sec tan^3 - 3 int sec tan^2 (1 + tan^2) dx
= sec tan^3 - 3 int tan d(sec) - 3 int tan^3 d(sec)

This should give you enough confidence to continue in this direction and get the answer.

haruspex