Weierstrass Substitution

[x(x-y)]

[latex]\displaystyle \int \frac{dx}{a\sin x + b\cos x + c\tan x} \hspace{10mm}a,b,c \in \mathbb{R} \ne 0[/latex]

 

[latex]let \; t = \tan \left(\frac{x}{2}\right) \Rightarrow dt = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right) dx[/latex]

 

[latex]\Rightarrow dx = \frac{2\, dt}{\sec^2\left(\frac{x}{2}\right)} = \frac{2\, dt}{1+ t^2}[/latex]

 

[latex]\sin \left(\frac{x}{2}\right) = \frac{t}{\sqrt{1 + t^2}}[/latex]

 

[latex]\cos \left(\frac{x}{2}\right) = \frac{1}{\sqrt{1 + t^2}}[/latex]

 

[latex]\Rightarrow \sin x = 2\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) = \frac{2t}{1 + t^2}[/latex]

 

[latex]\Rightarrow \cos x = \cos^2 \left(\frac{x}{2}\right) - \sin^2 \left(\frac{x}{2}\right) = \frac{1-t^2}{1+t^2}[/latex]

 

[latex]\Rightarrow \tan x = \frac{2\tan \left(\frac{x}{2}\right)}{1-\tan^2 \left(\frac{x}{2}\right)} = \frac{2t}{1-t^2}[/latex]

 

[latex]\Rightarrow \displaystyle \int \frac{\frac{2\, dt}{1 + t^2}}{\frac{2at}{1 +t^2} + \frac{b(1-t^2)}{1 + t^2} + \frac{2ct}{1-t^2}}[/latex]

 

[latex]\Rightarrow \displaystyle \int \frac{2\, dt}{2at + b(1-t^2) + \frac{2ct(1+t^2)}{1-t^2}}[/latex]

 

Where does one go from here? I multiplied numerator and denominator by 1-t^2 but that just left me with a quadratic over a quartic - I suppose I could try partial fractions but the quartic would be difficult to factorise.

 

P.S. This is not a homework problem, just me "doodling".

Edited January 20, 2013 by x(x-y)