Strategies for Solving Integrals Involving Trigonometric Functions

[_alexis_]

Hello.

I was trying to solve Lagrangian equation and I manage to reduce second order differential equation that I got:

\ddot{\varphi}+\alpha\frac{tan\varphi}{cos^{2}\varphi}=0;

where \alpha is a constant,

to first order differential equation:

\dot{\varphi}^{2}+ \alphatan^{2}\varphi -C=0;

where C is integration constant and from starting conditions I calculated it to be:
C= 8\alpha.


Now all I have left is this integral to solve:

\int \frac{d\varphi}{\sqrt{8-tan^{2}\varphi}}

But I can't find the right substitution.


I did try using some trigonometric identities to make this integral easier to solve or familiar but I didn't manage to get anywhere with it.

 

[MathematicalPhysicist]

The integral you want to calculate is \int\frac{d\phi}{\sqrt{1-tan^2(\phi)}} (obviously your integral we can factor out the 8 by suitable change of variables.

so use the fact that 1-tan^2(phi)= (cos^2()-sin^2())/cos^2()=cos(2phi)/cos^2()
after taking the sqrt and reciprocal you get:
cos(phi)dphi/sqrt(cos(2 phi)) = sqrt((cos(2 phi)+1)/(2cos(2 phi))) dphi
Now put cos(2 phi) = t to get an integral of the form (without the right factors obviously :-)):

-sqrt((t+1)/t) sqrt(1/(1-t^2) dt = sqrt(1/(t(1-t)) dt = sqrt(1/(-(t-1/2)^2+1/4)) dt

and that's look like a nice integral of the form:
ds/sqrt(1-s^2)

which its integral is?

 

[_alexis_]

Thank you for your answer.
You explained it very nicely.
And the solution for that integral is arcussin(s).

But...
I tried substituting \frac{tan\varphi}{\sqrt{8}}=tan\theta but I didn't get desired result because I got this:
d\varphi=\frac{\sqrt{8}}{1-4cos^{2}\theta}d\theta

But I may have made the mistake somewhere (and I made many stupid mistakes today).