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I figured out to use substitution $u=\tan\frac{x}{2}$ and arrived at integrate $\int_0^{\infty} \frac{2}{1+u^2+t(1-u^2)}du$ but am stuck here. Appreciate it if someone can drop some hint on how to proceed. Thank you.

Gary
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Anonymous123
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  • Now go from $u$ to $s$ via $$ u=\sqrt {\frac{{1 + t}}{{1 - t}}} s, $$ with $t\neq 1$. – Gary Oct 25 '21 at 11:33
  • Here is a similar question: https://math.stackexchange.com/questions/3657045/integral-of-trigonometric-function-with-parameter – Minus One-Twelfth Oct 25 '21 at 11:34
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    Presumably, $t<1$ to avoid poles. Then the denominator is proportional to $\frac{1-t}{1+t}u^2+1$ which reminds an $\arctan$. Rescale the variable. –  Oct 25 '21 at 11:42

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I am assuming you want to add the condition that $\lvert t\rvert<1$, otherwise the integral does not converge. Note that if $\lvert t\rvert<1$, then $1+t\cos x\neq0,$ so there is no issue of possibly having an improper integral.

We continue from what you found:

\begin{align*} \int_0^\infty\frac2{1+t+(1-t)u^2}\mathop{du}&=\frac{2}{1+t}\int_0^\infty\frac{1}{1+\left(\sqrt{\frac{1-t}{1+t}}u\right)^2}\mathop{du}\\ &=\frac{2}{1+t}\left.\left[\sqrt{\frac{1+t}{1-t}}\arctan\left(\sqrt{\frac{1-t}{1+t}}u\right)\right]\right|_0^\infty\\ &=\frac2{1+t}\cdot\frac{\sqrt{1-t^2}}{1-t}\cdot\left(\frac\pi2\right)\\ &=\frac{\pi}{\sqrt{1-t^2}} \end{align*}

Bonnaduck
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