Calculus, Integrals with Natural Logarithms

[ermac]

Homework Statement

∫tan^2(2x)/sec2x dx; u=sec2x; du=1/2tan^2(2x)dx.

Homework Equations

∫1/x(dx)-ln|x|+C.
∫1/u(du)=ln|u|+C

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.

 

[Char. Limit]

Homework Statement

∫tan^2(2x)/sec2x dx; u=sec2x; du=1/2tan^2(2x)dx.

Homework Equations

∫1/x(dx)-ln|x|+C.
∫1/u(du)=ln|u|+C

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.

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.

 

[ermac]

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

 

[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).

 

[ermac]

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

 

[Char. Limit]

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

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...

 

[ermac]

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...

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?

 

[Char. Limit]

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?

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.

 

[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)?

 

[Char. Limit]

Now you've got the right answer.

 

[ermac]

Thank you for the help.