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A Who can help evaluate this integral?

  1. Oct 10, 2016 #1
    I would like to evaluate the following integral: $$\int_0^{\phi}\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{V/v}}$$Is there anyone out there who can help? I've tried integral tables, Abramowitz and Stegun, and various substitutions, but with no success.
     
  2. jcsd
  3. Oct 10, 2016 #2

    fresh_42

    Staff: Mentor

    Playing with Wolfram always yields a denominator ##(\sin (\frac{\theta}{2}) + \cos (\frac{\theta}{2}))^n## and a nominator ##f(\sin \theta , \cos \theta)## depending on ##\frac{V}{v}##. Is there additional information on ##V/v##?

    I don't know whether this helps, but it might be a hint.
    I first thought ##\int \sec \theta d \theta = \ln |\sec \theta + \tan \theta|## could help, but this resulted in a nasty expression like ##\int \csc \theta \sec'(\theta) e^{-k \int \sec(\theta) d \theta} d \theta##.

    [Please feel free to delete this post, if you prefer to keep your question on the "unanswered" list.]

    Edit: ##n## and ##y## as power worked as well: http://www.wolframalpha.com/input/?i=f(x)=int((sec^2(x))/((sec+x+++tanx)^n))dx
     
  4. Oct 11, 2016 #3
    This wolframalpha result is perfect. It is exactly what I was looking for. The terms in the denominator simplify to ##\cos{\theta}##, so my definite integral becomes:$$\int_0^{\phi}\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{r}}=\frac{1}{r^2-1}\left[r-\frac{(r\sec{\phi}+\tan{\phi})}{(\sec{\phi}+\tan{\phi})^r}\right]$$where r = V\v.

    I differentiated the result to confirm that it is indeed correct. Thank you so much for your help. I still have no idea how to do the integration, but, oh well.

    Chet
     
    Last edited: Oct 11, 2016
  5. Oct 11, 2016 #4

    fresh_42

    Staff: Mentor

    This formula is far more beautiful than those wolframalpha output. Even the formula collection on Wikipedia with some dozens of trig integrals hasn't it.

     
  6. Oct 12, 2016 #5
    I figured out a method of doing the integration. It makes use of the following two identities:
    $$\sec{\theta}=\frac{1}{2}(\sec{\theta}+\tan{\theta})+\frac{1}{2}(\sec{\theta}-\tan{\theta})\tag{1}$$
    $$\sec^2{\theta}-\tan^2{\theta}=1$$
    or, equivalently,
    $$(\sec{\theta}-\tan{\theta})=(\sec{\theta}+\tan{\theta})^{-1}\tag{2}$$
    So, in the integral,
    $$\sec^2{\theta}=\sec{\theta}\left[\frac{1}{2}(\sec{\theta}+\tan{\theta})+\frac{1}{2}(\sec{\theta}-\tan{\theta})\right]=\frac{1}{2}(\sec^2{\theta}+\sec{\theta}\tan{\theta})+\frac{1}{2}(\sec^2{\theta}-\sec{\theta}\tan{\theta})$$
    If we substitute this into the integral, we obtain:$$\int{\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{r}}}=\frac{1}{2}\int{\frac{(\sec^2{\theta}+\sec{\theta}\tan{\theta})d\theta }{(\sec{\theta} + \tan{\theta})^{r}}}+\frac{1}{2}\int{\frac{(\sec^2{\theta}-\sec{\theta}\tan{\theta})d\theta }{(\sec{\theta} + \tan{\theta})^{r}}}$$But, if we substitute Eqn. 2 into the 2nd integral, we obtain:
    $$\int{\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{r}}}=\frac{1}{2}\int{\frac{(\sec^2{\theta}+\sec{\theta}\tan{\theta})d\theta }{(\sec{\theta} + \tan{\theta})^{r}}}+\frac{1}{2}\int{\frac{(\sec^2{\theta}-\sec{\theta}\tan{\theta})d\theta }{(\sec{\theta} - \tan{\theta})^{-r}}}$$Both integrands are exact differentials, and thus we immediately obtain:
    $$\int{\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{r}}}=\frac{1}{2}\left(\frac{(\sec{\theta} + \tan{\theta})^{1-r}}{(1-r)}+\frac{(\sec{\theta} - \tan{\theta})^{r+1}}{(r+1)}\right)+C$$Again making use of Eqn. 2 and rearranging yields:
    $$\int{\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{r}}}=\frac{1}{2}\left(\frac{(\sec{\theta} + \tan{\theta})}{(1-r)(\sec{\theta} + \tan{\theta})^r}+\frac{(\sec{\theta} - \tan{\theta})}{(1+r)(\sec{\theta} + \tan{\theta})^r}\right)+C$$
     
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