# Integration... wtf am I missing here?

Registered User regular
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

• Registered User regular
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
• Registered User regular
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.

• Registered User regular
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
• Registered User regular
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.