# Math Help - Help with a integration problem (trigonometric substitution)

1. ## Help with a integration problem (trigonometric substitution)

I'm having a test on integration on Monday and I'm having some major problems with solving integrations involving trigonometric substitution. There's a sample of a problem I'm having issues with:
Code:
I = Integration symbol
u = dummy variable for substitution

I tan^2(6x) * dx

My work:

sin^2(6x)
I   ----------  * dx         u = cos(6x)
cos^2(6x)                du = -6sin(6x) * dx

du
dx =   ------------
-6sin(6x)

sin^2(6x)        du
I   ---------- *   --------
u          -6sin(6x)

-1     sin(6x)
--- I  --------  * du
6         u
This is where I get stuck. I remember my teacher saying that I have to get everything in terms of u before I go substituting u back into the equation, however, I don't see a feasible way to do this. If someone could step my through this one so I can see how it's done, I'd really appreciate it.

If you all need any more information about the problem or something I'll do my best to provide. Regards.

2. Originally Posted by Gamerdude
I'm having a test on integration on Monday and I'm having some major problems with solving integrations involving trigonometric substitution. There's a sample of a problem I'm having issues with:
Code:
I = Integration symbol
u = dummy variable for substitution

I tan^2(6x) * dx

My work:

sin^2(6x)
I   ----------  * dx         u = cos(6x)
cos^2(6x)                du = -6sin(6x) * dx

du
dx =   ------------
-6sin(6x)

sin^2(6x)        du
I   ---------- *   --------
u          -6sin(6x)

-1     sin(6x)
--- I  --------  * du
6         u
This is where I get stuck. I remember my teacher saying that I have to get everything in terms of u before I go substituting u back into the equation, however, I don't see a feasible way to do this. If someone could step my through this one so I can see how it's done, I'd really appreciate it.

If you all need any more information about the problem or something I'll do my best to provide. Regards.
You need to cancel the common sin(6x) first.

Then
u = cos(6x)

So
u^2 = cos^2(6x) = 1 - sin^2(6x)

sin^2(6x) = 1 - u^2

and
sin(6x) = (+/-)sqrt{1 - u^2}

You'll need to look at the limits of integration to decide where you use the + and - signs.

-Dan

3. Thanks a ton. I'm going to do some more practice problems in a few hours and see how I do. If I run into trouble again I'll post back here. Thanks for such a quick response.

4. Hello, Gamerdude!

tan²(6x) dx

We have: .
[sec²(6x) - 1] dx .= .(1/6)tan(6x) - x + C

5. Thanks a lot, I've been able to solve most of the problems I've been having trouble with over the past few days. My headache is now basically gone . I'm only stuck on two problems (which implement the same concept), here they are:
Code:
∫csc^2(3x) * cot(3x) dx
and
Code:
∫tan(2x) * sec^2(2x) dx

I know that these two problems are essentially the same as far as the answering them goes. I found a integration calculator online which shows me the answer to the problems, but unfortunately it doesn't show the steps. Could someone walk me through one of these? If I can do one I should be able to do the other. Thanks again all.

6. Hello, Gamerdude!

You seem to be making hard work out of these.
. . Just pick an appropriate substitution.

csc²(3x)·cot(3x) dx

Let u = cot(3x) . . . . du = -3·csc²(3x) dx

The integral becomes: .(-1/3)
u du

tan(2x)·sec²(2x) dx

Let u = tan(2x) . . . . du = 2·sec²(2x) dx

The integral becomes: .(1/2)
u du

7. Soroban, I think where I'm getting lost is that once you have
Code:
(1/2) ∫ u du
for: ∫ tan(2x)·sec²(2x) dx

and you plug u and du back into the equation, you end up with almost the same function as when you started. I know this is where I'm not getting it, but I don't know why.

So, in the end, after I plug u and du back in, this is what I always come up with:
Code:
(1/2)∫ tan(2x)·2sec²(2x)
I know this isn't the answer (The answer should be getting is (1/4) * sec²(2x) from what I've gathered). What am I doing wrong?

8. Originally Posted by Gamerdude
Soroban, I think where I'm getting lost is that once you have
Code:
(1/2) ∫ u du
for: ∫ tan(2x)·sec²(2x) dx

and you plug u and du back into the equation, you end up with almost the same function as when you started. I know this is where I'm not getting it, but I don't know why.

So, in the end, after I plug u and du back in, this is what I always come up with:
Code:
(1/2)∫ tan(2x)·2sec²(2x)
I know this isn't the answer (The answer should be getting is (1/4) * sec²(2x) from what I've gathered). What am I doing wrong?
remember, du = 2sec^2(2x) dx, so plugging in u and du back into the formula we would get exactly the one we started with. however, that is not what you should do. you should integrate with respect to u first, and then plug in the value for u. try it

9. So, if I integrate with respect to u, I will get
Code:
(1/4) * tan²(2x) + c
right? Is this my final answer or do I have to do something else?

10. Originally Posted by Gamerdude
So, if I integrate with respect to u, I will get
Code:
(1/4) * tan²(2x) + c
right? Is this my final answer or do I have to do something else?
that's the final answer, if you were doing the definate integral it wouldn't be

11. Awesome. I think I finally get it now. Just to clarify my understanding, the following integration:
Code:
∫csc²(3x) * cot(3x) dx
will come out to equal:
Code:
(-1/6) * cot²(3x)
Correct?

12. Originally Posted by Gamerdude
Awesome. I think I finally get it now. Just to clarify my understanding, the following integration:
Code:
∫csc²(3x) * cot(3x) dx
will come out to equal:
Code:
(-1/6) * cot²(3x)
Correct?
Correct

13. Awesome! Thank you everyone that helped me out. I'm feeling much more confident about my test tomorrow. If I run into any last minute issues I'll be sure to post back. Positive rep points to all!

14. Originally Posted by Gamerdude
Awesome! Thank you everyone that helped me out. I'm feeling much more confident about my test tomorrow. If I run into any last minute issues I'll be sure to post back. Positive rep points to all!
great! keep practicing. and good luck!