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Jimmy King
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.

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.

0

## Posts

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

Instead! let u = 3x

(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.

enlightenedbumonI'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.

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.