Question

prove intergral

Answer 1

\[\int\limits _0^1\frac{1}{\alpha-\cos x}dx=\frac{\pi}{\sqrt{\alpha^2-1}}\]

Answer 2

i tried \[\tan \frac{x}{2}=u\]
i have
\[\huge \int\limits \frac{1}{\alpha-\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=2\huge \int\limits \frac{1}{\alpha-1+(1+\alpha)t^2}dt\]

Answer 3

\[\huge \frac{2}{\alpha+1}\int\limits \frac{dt}{(\sqrt{\frac{\alpha-1}{\alpha+1}})^2+t^2}=\frac{2}{\alpha+1}\tan^{-1}(\frac{t}{\sqrt{\frac{\alpha-1}{\alpha+1}}})\]

Answer 4

first of all tan pi/2 is undefined,is there anything wrong here?

Answer 5

oh something missing
the whole thing shud be divided by
\[\sqrt{\frac{\alpha-1}{\alpha+1}}\]

Answer 6

\[\huge \frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\frac{\frac{x}{2}}{\sqrt{\frac{\alpha-1}{\alpha+1}}}\]

Answer 7

Did you mutliply by conjugate?

Answer 8

yes

Answer 9

Wait, you're proving. My bad, I thought you were solving using trig sub.