Results 1 to 6 of 6

Math Help - Antiderivative help, please.

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    53

    Antiderivative help, please.

    Hello,

    I am stuck with this particular integral:

    ∫ tan^4(2x)sec^4(2x)dx

    As far as I could go was to:

    ∫ (tan^2(2x)sec^2(2x))^2 dx

    Any help is much appreciated,

    Dranalion.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Feb 2009
    From
    Posts
    79
    Quote Originally Posted by Dranalion View Post
    Hello,

    I am stuck with this particular integral:

    ∫ tan^4(2x)sec^4(2x)dx

    As far as I could go was to:

    ∫ (tan^2(2x)sec^2(2x))^2 dx

    Any help is much appreciated,

    Dranalion.
    Suggestion
    \int (tan^{2}(2x)sec^2(2x))^{2} dx=
    \int (tan^{2}(2x) ( 1+tan^{2}(2x) )^{2} dx
    \int (tan^{4}(2x)+tan^{2}(2x) )^{2} dx

    Multiply out
    \int (tan^{8}(2x)+2tan^{6}(2x)  + tan^{4}(2x)) dx


    Use the reduction formula for each of the powers of tanx.

    There might be a cleaner way to do this though.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Dranalion View Post
    Hello,

    I am stuck with this particular integral:

    ∫ tan^4(2x)sec^4(2x)dx

    As far as I could go was to:

    ∫ (tan^2(2x)sec^2(2x))^2 dx

    Any help is much appreciated,

    Dranalion.
    Hint:

    \begin{aligned}\int\tan^4\!\left(2x\right)\sec^4\!  \left(2x\right)\,dx & = \int\tan^4\!\left(2x\right)\sec^2\!\left(2x\right)  \sec^2\!\left(2x\right)\,dx\\ & = \int\tan^4\!\left(2x\right)\left(\tan^2\!\left(2x\  right)+1\right)\sec^2\!\left(2x\right)\,dx\\ & = \int\left(\tan^6\!\left(2x\right)+\tan^4\!\left(2x  \right)\right)\sec^2\!\left(2x\right)\,dx\end{alig  ned}

    Now make the substitution u=\tan\!\left(2x\right)

    Can you continue?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Feb 2009
    From
    Posts
    79
    The above solution is cleaner than what i posted.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2009
    Posts
    53
    I understand the process you have used to find an easier integral to evaluate, and also, from the textbook, I know the answer is going to be:

    [5*tan^7(2x) + 7*tan^5(2x)] / 70

    However, I do not understand the method used to get there!

    Can anyone please assist me here?!


    Any help much appreciated, again,

    Dranalion.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Feb 2009
    From
    Posts
    79
    You have to remember that the answer in your book is simplified.

    If you actually evaluate the integral you should get a similar answer which you could simplify; and they would end up been exactly equivalent.

    If you don't get the same answer it's alright as long as the logic is correct and there are no mistakes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. antiderivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 20th 2009, 11:04 AM
  2. Antiderivative Help!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 16th 2009, 03:40 PM
  3. Antiderivative of 2^t
    Posted in the Calculus Forum
    Replies: 11
    Last Post: April 6th 2009, 07:00 PM
  4. Antiderivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 4th 2009, 05:31 AM
  5. Antiderivative (sin x)(ln x)
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 6th 2008, 08:31 PM

Search Tags


/mathhelpforum @mathhelpforum