Results 1 to 3 of 3
Like Tree3Thanks
  • 3 Post By ReneG

Math Help - Are these two integral answers equivalent?

  1. #1
    Junior Member
    Joined
    Nov 2010
    From
    Colorado
    Posts
    42
    Thanks
    2

    Are these two integral answers equivalent?

    \int sec^2xtanxdx
    I let u=tanx
    du=sec^2xdx
    \int udu
    \frac{u^2}{2} + C

    \frac{1}{2}tan^2x + C

    The answer key let u=sec^2x
    Their answer is
    \frac{1}{2}sec^2x + C

    It doesn't seem that they could be the same. What did I do wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member ReneG's Avatar
    Joined
    Mar 2013
    From
    United States
    Posts
    66
    Thanks
    9

    Re: Are these two integral answers equivalent?

    Both substitutions are correct, the book's substitution would have been easier though.

    Your answer was \frac{1}{2} \tan^2x + \mathrm{C}

    Although it really isn't intuitive, from there you can use the tangent identity \tan^2\theta = \sec^2\theta - 1 to rewrite your integral as \frac{\sec^2x}{2} - \frac{1}{2} + \mathrm{C} now -1/2 is a constant and any constant plus an arbitrary constant is still an arbitrary constant so your answer becomes \frac{\sec^2x}{2} + \mathrm{C}

    When all else fails, you can always try writing those trig functions in terms of sine and cosine. For this problem it would be
    \int \frac{\sin x}{\cos x} \frac{1}{ \cos^2x}\,\,dx = \int\frac{\sin x}{\cos^3x}\,\,dx \,\,\,\,\, u =\cos x \rightarrow dx = -\sin x \,\,du
    -\int \frac{1}{u^3}\,\,du = -\frac{-1}{2u^2} + \mathrm{C} = \frac{1}{2} \frac{1}{\cos^2x} + \mathrm{C}=  \frac{\sec^2x}{2} + \mathrm{C}
    that's a lot more unnecessary work, but it's a surefire way to get an answer
    Last edited by ReneG; June 30th 2013 at 09:57 AM.
    Thanks from Shakarri, wondering and topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2010
    From
    Colorado
    Posts
    42
    Thanks
    2

    Re: Are these two integral answers equivalent?

    Awesome, thank you for the explanation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 10th 2012, 11:30 PM
  2. Incorrect Indefinite Integral Answers
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 30th 2011, 04:31 AM
  3. [SOLVED] Equivalent Integral for restricted values
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 13th 2011, 10:33 AM
  4. Polar equivalent integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 13th 2009, 06:22 AM
  5. answers for the following integral questions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 29th 2007, 04:32 AM

Search Tags


/mathhelpforum @mathhelpforum