Evaluate $\int \frac{dx}{\sqrt{3} - \cos x - \sin x} $
Answer:
$\int \frac{dx}{\sqrt{3} - \cos x - \sin x} $
$=\int \frac{dx}{\sqrt{3} - \sqrt{2}(\sin (x + \frac{\pi}{4})} $
Substitute $u = x + \frac{\pi}{4}$, and we have
$\int \frac{dx}{\sqrt{3} - \sqrt{2}(\sin (x + \frac{\pi}{4})} $
$=\int \frac{du}{\sqrt{3} - \sqrt{2}\sin u} $
We will derive a formula for $\int \frac{dx}{a - b\sin x} $
Substitute $ \sin x = \frac{2t}{1+t^2}$, then
$\cos x = \frac{1-t^2}{1+t^2} $ and $tan x = \frac{2t}{1-t^2}$, and also
$\cos x \; dx = \frac{2(1-t^2)}{(1+t^2)^2} \; dt$
or $dx = \frac{2}{1 + t^2} $
or,
So, $\int \frac{dx}{a - b\sin x} $
$= \int \frac{ \frac{2}{1 + t^2}\;dt } { a - b \frac{2t}{1+t^2} } $
$= \int \frac{2 \; dt}{a(1+t^2) - 2bt} $
$= \frac{2}{a} \int \frac{dt}{(t - \frac{b}{a})^2 + 1 - \frac{b^2}{a^2}} $
Substituting $s = t - \frac{b}{a}$, we have
$\frac{2}{a} \int \frac{dt}{(t - \frac{b}{a})^2 + 1 - \frac{b^2}{a^2}} $
$= \frac{2}{a} \int \frac{ds}{s^2 +\frac{a^2-b^2}{a^2}} $
$= \frac{2}{a} \frac{a}{\sqrt{a^2 - b^2}} \tan^{-1} \frac{as}{\sqrt{a^2 - b^2}} + C$
$= \frac{2}{\sqrt{a^2 - b^2}} \tan^{-1} \frac{at - b }{\sqrt{a^2 - b^2}} + C$
$= \frac{2}{\sqrt{a^2 - b^2}} \tan^{-1} \frac{a\tan \frac{x}{2} - b }{\sqrt{a^2 - b^2}} + C$
So,
$\int \frac{du}{\sqrt{3} - \sqrt{2}\sin u} $
$= 2 \tan^{-1} \left( \sqrt{3}\tan \frac{u}{2} - \sqrt{2} \right) + C$
$= 2 \tan^{-1} \left( \sqrt{3}\tan \frac{x + \frac{\pi}{4}}{2} - \sqrt{2} \right) + C$
Solution was inspired by +Snail Erato work on the integral of the following
$\int \frac{dx}{\cos^3 x + \sin^3 x} $
here in two these pages:
1. Page 1
2. Page 2
