Tricky Calculus Question II

The substitution $t = \tan\tfrac12x$ gives $\begin{aligned} I \; = \; \int_{-\tfrac12\pi}^{\tfrac12\pi} \frac{dx}{1+ \cos a \sin x} & = \; \int_{-1}^1 \frac{1}{1 + \cos a \frac{2t}{1+t^2}} \frac{2\,dt}{1+t^2} \; = \; 2\int_{-1}^1 \frac{dt}{1 + 2\cos a t + t^2}\\ & = \; 2\int_{-1}^1 \frac{1}{(t+\cos a)^2 + \sin^2a}\,dt \; = \; \frac{2}{|\sin a|}\Big[\tan^{-1}\frac{t + \cos a}{|\sin a|}\Big]_{-1}^1 \\ & = \; \frac{2}{|\sin a|}\Big(\tan^{-1}\big(\cot\tfrac12a\big) + \tan^{-1}\big(\tan\tfrac12a\big)\Big) \end{aligned}$ Just to keep things simple, let us consider this integral for $0 < a < \tfrac12\pi$ . Then $I \; = \; \frac{2}{\sin a}\big(\tfrac12\pi - \tfrac12a + \tfrac12a\big) \; = \; \frac{\pi}{\sin a}$ which makes the answer $k = \boxed{1}$ .

$I= \int_{\frac {-\pi}{2}}^{\frac {\pi}{2}} \frac {dx}{1+\cos a\sin x}=\int_{0}^{\frac {\pi}{2}} \frac {dx}{1+\cos a\sin x}+\int_{\frac {-\pi}{2}}^{0} \frac {dx}{1+\cos a\sin x}=\int_{0}^{\frac {\pi}{2}} \frac {2dx}{1-\cos ^2 a\sin ^2 x}=\int_0^{\frac {\pi}{2}} \frac {2\csc ^2 x dx}{\cot^2 x +1-\cos^2 a}$

Substituting $\cot x=u$ we get $I=\int_0^{\infty} \frac {2du}{u^2+ \sin^2 a}= \frac {\pi}{\sin a}$

Hence $k=1$