Calculus, Integrals with Natural Logarithms

1. Apr 15, 2010

ermac

1. The problem statement, all variables and given/known data

∫tan^2(2x)/sec2x dx; u=sec2x; du=1/2tan^2(2x)dx.
2. Relevant equations
∫1/x(dx)-ln|x|+C.
∫1/u(du)=ln|u|+C

3. The attempt at a solution
This is me trying to rewrite the equation. (sin^2(2x)/cos^2(2x))/(1/cos2x), (sin^2(2x))/(cos(2x)).

Honestly, I feel lost trying to find a differential on the denominator, to change into the numerator.

2. Apr 15, 2010

Char. Limit

But you as good as had it after the substitution... you just need to use u=sec(2x), du=2tan^2(2x) dx... which you had, except for the fact that you flipped the two. Don't try to do all of what you're doing, when you had it at the substitution. Now you just need to substitute.

3. Apr 15, 2010

ermac

I'm sorry, but what two did I flip?

4. Apr 15, 2010

Char. Limit

The two. As in you flipped the 2, in your du expression.

Instead of .5tan^2(2x), you should have 2tan^2(2x).

5. Apr 15, 2010

ermac

So, no changing into different trig functions or anything like that? Just ln|sec(2x)|+C?

6. Apr 15, 2010

Char. Limit

No, no changing into different trig functions, but you did forget that two.

If du=tan^2(2x) dx, then you'd be right.

But du=2tan^2(2x) dx.

That two needs to be accounted for...

7. Apr 15, 2010

ermac

Sorry about my guess and check, but would that two essentially come out to the front because of the chain rule, making it 2ln|sec(2x)|+C?

8. Apr 15, 2010

Char. Limit

Almost. You have tan^2(2x) dx in the integrand, so you want to isolate it in the substitution formula by dividing both sides by two.

du=2tan^2(2x) dx

(1/2)du=tan^2(2x) dx

Then you put the 1/2 du in, and take the constant out of the integrand.

9. Apr 15, 2010

ermac

Would the constant in this case be 2x? Oh wait, could 1/2ln|sec(2x)|+C, remove the 2 in front of the tan^2(2x)?

Last edited: Apr 15, 2010
10. Apr 15, 2010

Char. Limit

Now you've got the right answer.

11. Apr 15, 2010

ermac

Thank you for the help.