Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

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February 17th 2013, 11:44 AM
UBCBOY

Evaluate 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!
February 17th 2013, 11:52 AM
ILikeSerena

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by UBCBOY

Hi, my question is:

Evaluate the indefinite integral:
integral cos(√x)/√x dx

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

Hi UBCBOY! :)

Integration by parts would work... why don't you try it?

Alternatively you can try the substitution u=√x.

Generally substitution and integration by parts are the way to go to integrate a product.
February 17th 2013, 11:56 AM
UBCBOY

Re: 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.
February 17th 2013, 12:01 PM
SlipEternal

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Just as ILikeSerena said, integration by parts and substitution are you friends. Try $u=2x$ as your substitution.

You added a possible substitution. $u=\sec(2x), du=2\sec(2x)\tan(2x)$ works. Then, you are integrating only $-4du$ .
February 17th 2013, 12:01 PM
ILikeSerena

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by UBCBOY

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!

Quote:

Originally Posted by UBCBOY

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.

It seems you already have a good idea.
What do you get if you do the substitution?
February 17th 2013, 12:10 PM
UBCBOY

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by ILikeSerena

It seems you already have a good idea.
What do you get if you do the substitution?

I'm just really confused and I don't know if I'm doing it right, because if I let u = sec(2x), then du = sec(2x)tan(2x)dx am I right?

And if I sub that in, doesn't the sec(2x)tan(2x) just get canceled out if I sub out dx for du/sec(2x)tan(2x)...

:S :S :S
February 17th 2013, 12:14 PM
ILikeSerena

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by UBCBOY

I'm just really confused and I don't know if I'm doing it right, because if I let u = sec(2x), then du = sec(2x)tan(2x)dx am I right?

And if I sub that in, doesn't the sec(2x)tan(2x) just get canceled out if I sub out dx for du/sec(2x)tan(2x)...

:S :S :S

It should be: du = 2 sec(2x)tan(2x)dx

And yes, most of your integral gets canceled out, but not all.
Isn't that a good thing?
What's left looks to be easy to integrate. :cool:
February 17th 2013, 12:28 PM
UBCBOY

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by ILikeSerena

It seems you already have a good idea.
What do you get if you do the substitution?

Quote:

Originally Posted by ILikeSerena

It should be: du = 2 sec(2x)tan(2x)dx

And yes, most of your integral gets canceled out, but not all.
Isn't that a good thing?
What's left looks to be easy to integrate. :cool:

Could you show me your work? I'm not getting the right answer :(
February 17th 2013, 12:32 PM
ILikeSerena

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by UBCBOY

Could you show me your work? I'm not getting the right answer

Can you show what you've got?
February 17th 2013, 12:42 PM
UBCBOY

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by ILikeSerena

Can you show what you've got?

After substituting u = sec(2x), I get:

-8∫u*tan(2x) du/(2sec(2x)tan(2x)

So the tan(2x) cancels out, and I'm left with...

-8∫u/2sec(2x) and I don't know what to do here, since I can't sub back in sec(2x) for u until I integrate the u right?
February 17th 2013, 12:51 PM
ILikeSerena

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by UBCBOY

After substituting u = sec(2x), I get:

-8∫u*tan(2x) du/(2sec(2x)tan(2x)

So the tan(2x) cancels out, and I'm left with...

-8∫u/2sec(2x) and I don't know what to do here, since I can't sub back in sec(2x) for u until I integrate the u right?

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.
February 17th 2013, 01:20 PM
UBCBOY

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by ILikeSerena

It seems you already have a good idea.
What do you get if you do the substitution?

Quote:

Originally Posted by ILikeSerena

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.

So if I plug in sec(2x) for x, I am just left with -8∫1/2, which becomes -4x + C right...?

(It's not the right answer :()
February 17th 2013, 01:22 PM
ILikeSerena

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by UBCBOY

So if I plug in sec(2x) for x, I am just left with -8∫1/2, which becomes -4x + C right...?

(It's not the right answer :()

You should get -4u + C.
Now you can substitute u=sec(2x) back again.
February 17th 2013, 01:26 PM
UBCBOY

Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

Quote:

Originally Posted by ILikeSerena

It seems you already have a good idea.
What do you get if you do the substitution?

Quote:

Originally Posted by ILikeSerena

You should get -4u + C.
Now you can substitute u=sec(2x) back again.

That makes so much more sense...thank you! :)

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