Hi, my question is:

Evaluate the indefinite integral:

∫−8sec(2x)tan(2x)dx

Am I supposed to use substitution and let u = sec2x, and let du = sec(2x)tan(2x) dx?

I'm not sure how to approach this...thank you!

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- Feb 17th 2013, 11:44 AMUBCBOYEvaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
Hi, my question is:

Evaluate the indefinite integral:

∫−8sec(2x)tan(2x)dx

Am I supposed to use substitution and let u = sec2x, and let du = sec(2x)tan(2x) dx?

I'm not sure how to approach this...thank you! - Feb 17th 2013, 11:52 AMILikeSerenaRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 11:56 AMUBCBOYRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
Ah thanks for answering ILikeSerena! Coincidentally I figured how to approach that question literally a few seconds right after I had posted it, and decided to change my thread regarding another question I was having trouble with. Would you be able to help me with the new question?

Sorry for changing it. - Feb 17th 2013, 12:01 PMSlipEternalRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
Just as ILikeSerena said, integration by parts and substitution are you friends. Try $\displaystyle u=2x$ as your substitution.

You added a possible substitution. $\displaystyle u=\sec(2x), du=2\sec(2x)\tan(2x)$ works. Then, you are integrating only $\displaystyle -4du$. - Feb 17th 2013, 12:01 PMILikeSerenaRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 12:10 PMUBCBOYRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 12:14 PMILikeSerenaRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 12:28 PMUBCBOYRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 12:32 PMILikeSerenaRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 12:42 PMUBCBOYRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 12:51 PMILikeSerenaRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
You now have u and x together in the same integral expression.

That is not supposed to happen.

Since you're switching to u, you should replace sec(2x) with an expression that contains u.

Do you have such an expression?

Oh wait! You have u=sec(2x).

So yes, you should back sub that in to get an expression that is only in u. - Feb 17th 2013, 01:20 PMUBCBOYRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 01:22 PMILikeSerenaRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx
- Feb 17th 2013, 01:26 PMUBCBOYRe: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx