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:
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.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
If you all need any more information about the problem or something I'll do my best to provide. Regards.
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:
andCode:∫csc^2(3x) * cot(3x) dx
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.
You seem to be making hard work out of these.
. . Just pick an appropriate substitution.
∫ cscē(3x)·cot(3x) dx
Let u = cot(3x) . . → . . du = -3·cscē(3x) dx
The integral becomes: .(-1/3) ∫ u du
∫ tan(2x)·secē(2x) dx
Let u = tan(2x) . . → . . du = 2·secē(2x) dx
The integral becomes: .(1/2) ∫ u du
Soroban, I think where I'm getting lost is that once you haveCode:(1/2) ∫ u du for: ∫ tan(2x)·secē(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:
I know this isn't the answer (The answer should be getting is (1/4) * secē(2x) from what I've gathered). What am I doing wrong?Code:(1/2)∫ tan(2x)·2secē(2x)