# Integration... wtf am I missing here?

Ok, I get integration, normally get these right, etc. I'm doing some review before a test over it tonight and I have one problem that I don't understand how the teacher came to. Wolfram Alpha get the same answer and I can even see the steps it took and it still doesn't make any sense.

You can see it here in a more readable format.
Integrate[sin3x/(2-2sin^2(3x)) dx]

The answer, apparently, is (1/6)sec(3x) +c

I can get it to (1/6)Integrate[tan(3x)] but the antiderivative of tan(x) is not sec(x)

Wolfram and I do this in a bit different order, but it should not matter.
The wolfram alpha steps then use u-du with u=3x and du=3dx (I pull out 1/2 to get sin/cos first, which becomes tan).
THen wolfram alpha does like me and pulls out the 1/2... and gets tan(u)sec(u) using some sorcery that my feeble mind can't comprehend. I, on the other hand, just have tan(u) and of course the 1/6 from pulling out 1/2 and the u-du stuff.

If you want to see the wolfram steps use this link and paste in the below.
sin(3x)/(2-2sin^2(3x))

What am I missing? My way is clearly wrong because tan(u) has no antiderivative, but I don't see where this sec(u) is coming from.

## Posts

• edited November 2012
OK, so:

Same as integrate (1/2) * sin (3x) / (1 - sin^2(3x)) dx
which is int (1/2) * sin (3x) / (cos^2(3x)) dx
which is (1/2) * sin (3x) / cos (3x) * 1/ cos (3x) dx
which is (1/2) tan (3x) * sec (3x) dx ... which is a dumb way to put it

(1/6) * sin u / cos^2 (u) du
Let w = cos u, dw = -sin u du

So we get:

(-1/6) * int (1/w^2) dw
= (-1/6) * -1/w + C
= (1/6) * 1/cos 3x + C
= (1/6) sec 3x + C

The important thing here is creative usage of u substitution. When you see the two standard trig functions in a fraction, it's generally a sign they want you to do this.

enlightenedbum on
Herbert Hoover got 40% of the vote in 1932. Friendly reminder.
Warren 2020
• Oh shit. I was doing 1-sin^2 = cos instead of cos^2. blargh. that's why I couldn't see how the hell a sec got in there, all I had was sin/cos instead of the proper (sin/cos)(1/cos).

I'll keep the double substitution trick in mind as well. I don't think he's going to do that to us on this test, though.

Thanks for the help.

• Really you could do both at once, but when you're learning this stuff for the first time it's probably best to just keep things simple and do one thing at a time.

For reference:
u = cos 3x
du = -3 * sin 3x dx

I mean, it's not super difficult, but simple is good when you're starting.

Herbert Hoover got 40% of the vote in 1932. Friendly reminder.
Warren 2020
• yeah, and I like to do small steps because then usually I catch the stupid stuff like I did here where I just flat out write it down wrong and keep going. Plus it makes my current teacher give me more points for showing work because he can see that I understood the concept and just did something dumb when it comes to graded work.