# A Who can help evaluate this integral?

1. Oct 10, 2016

### Staff: Mentor

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. Oct 10, 2016

### 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

3. Oct 11, 2016

### Staff: Mentor

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
4. Oct 11, 2016

### 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.

5. Oct 12, 2016

### Staff: Mentor

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:
$$\int{\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{r}}}=\frac{1}{2}\left(\frac{(\sec{\theta} + \tan{\theta})}{(1-r)(\sec{\theta} + \tan{\theta})^r}+\frac{(\sec{\theta} - \tan{\theta})}{(1+r)(\sec{\theta} + \tan{\theta})^r}\right)+C$$