Question

Question abut integral 3sec(3x)tan(3x)dx

Answer 1

In my notes I had this question and it had noted that I only needed to use du when doing a u sub.

u = sec(3x)
1/3du = sec(3x)tan(3x)dx

\[1/3 \int\limits du\]

Answer 2

\[1/3 * 3\sec(3x) + c\] = sec(3x) + c

Answer 3

now on wolfram http://www.wolframalpha.com/input/?i=integral+3sec%283x%29tan%283x%29dx

Answer 4

integral of sec3xtan3x=sec3x

Answer 5

they do a double u-sub though it's the same way.... Are there certain times we should use a double u-sub....

Answer 6

and also it seems that only using du is allowed, but why do we get rid of the u? I thought both are needed....

Answer 7

answer is sec3x

Answer 8

I know what the answer is, I already said it, why do you keep saying this?

Answer 9

I am not sure if I understand your question @KonradZuse, do you have a problem with the substitution itself?

Answer 10

I know how to solve it, what I was asking is it okay to just use du when we are u-subbing... Since this case only uses du, so I would assume yes... Then I was asking about wolfram's answer, it does 2 u-sub's so I was curious when should we use double u-subs?

Answer 11

ah now I understand, let me check wolf's answer first.

Answer 12

Thank you sir.

Answer 13

Well Wolfram only uses two substitutions because these are chained functions, the inside of the functions are 3x, so when you integrate that you have to take care of that inside function too.

So they use an additional substitution for that, but usually, most people don't do that, we just apply what's called an 'advanced' guess. It is obvious that it needs the 1/3 in front so if you derive it, it will cancel again.

Answer 14

In this example, you don't need that additional step because the derivative with one single u substitution matches the integral perfectly already.

Answer 15

If the inside function would be something of the form x^2, you would have to make a substitution for this first, so you can get it in a more suitable form.

Answer 16

hmm? you're saying sec(x^2)tan(x^2)dx?

Answer 17

so what I did above is correct, and you can use du and disregard u? WE MUST at least have du though?

Answer 18

I'm just curious about when we would need to use a double sub...

Answer 19

I think I used it once this semester.

Answer 20

You will always need minimum one substitution for this kind of problems, unfortunately it's a bit hard for me to think about one myself at the top of my head where you need two substitution.

But try to think about it like this, the substitution method is nothing like a reverse of the chain rule.

The chain rule says this (verbally): Take the derivative of the outside (function) with respect to the inside and multiply that by the derivative of the inside function.

So if you reverse that statement (a bit hard to read it from the end to the start) but it will say something like: find a respective term for the inside function first, then take care of the outside function.

Answer 21

yeah I've been told that too... "But try to think about it like this, the substitution method is nothing like a reverse of the chain rule."

Answer 22

So like I said before.... so what I did above is correct, and you can use du and disregard u? WE MUST at least have du though?

Answer 23

yes, exactly

\[\large f(x)=3\sec(3x)\tan(3x) \]
Let
\[ \large u= \sec(3x) \longrightarrow \frac{du}{dx}=3\sec(3x)\tan(3x)=f(x)\]
So here one substitution is enough and as you said, we must at least have du

\[\Large \int f(x)dx= \int \frac{du}{dx}dx = \int 1 du \]

Answer 24

Okay cool cool.... Interesting how we can just get rid of u tho....

Answer 25

Yes in this case it works out, but it's still there, in the derivative itself, so it looks a bit disguised.

Answer 26

\[\large u= \sec(3x) \longrightarrow \frac{du}{dx}=3\underbrace{\sec(3x)}_u\tan(3x)=f(x) \]

Answer 27

yeah well we do get u when we solve I guess.... So it does have a role...