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

  1. #1
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    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!
    Last edited by UBCBOY; February 17th 2013 at 11:49 AM.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by UBCBOY View Post
    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.
    Last edited by ILikeSerena; February 17th 2013 at 12:04 PM.
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    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.
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    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.
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  5. #5
    Super Member ILikeSerena's Avatar
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by UBCBOY View Post
    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 View Post
    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?
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by ILikeSerena View Post
    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
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  7. #7
    Super Member ILikeSerena's Avatar
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by UBCBOY View Post
    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.
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by ILikeSerena View Post
    It seems you already have a good idea.
    What do you get if you do the substitution?
    Quote Originally Posted by ILikeSerena View Post
    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.
    Could you show me your work? I'm not getting the right answer
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  9. #9
    Super Member ILikeSerena's Avatar
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by UBCBOY View Post
    Could you show me your work? I'm not getting the right answer
    Can you show what you've got?
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by ILikeSerena View Post
    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?
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  11. #11
    Super Member ILikeSerena's Avatar
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by UBCBOY View Post
    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.
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by ILikeSerena View Post
    It seems you already have a good idea.
    What do you get if you do the substitution?
    Quote Originally Posted by ILikeSerena View Post
    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 )
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  13. #13
    Super Member ILikeSerena's Avatar
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by UBCBOY View Post
    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.
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    Re: Evaluate the indefinite integral ∫−8sec(2x)tan(2x)dx

    Quote Originally Posted by ILikeSerena View Post
    It seems you already have a good idea.
    What do you get if you do the substitution?
    Quote Originally Posted by ILikeSerena View Post
    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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