Show that $\int_0^{\pi/2} \frac {\sin(u+a\tan u)} {\sin u}\,\mathrm d u=\pi/2$ I am trying to show that $$\int_0^{\pi/2} \frac {\sin(u+a\tan u)} {\sin u}\,\mathrm d u=\frac {\pi} 2$$
Can this be done via contour integration? I'm not really sure which contour to pick. I have tried substitutions like $\pi/2 - u$ but they haven't helped. I have tried differentiating with respect to $a$ too. I got 
$I'(a)=\int_0^{\pi/2} \frac {\cos(u+a\tan u)} {\cos u}\,\mathrm d u$
And I know 
$I(a)=\int_0^{\pi/2} \frac {\cos(u-a\cot u)} {\cos u}\,\mathrm d u$
This came up on an undergraduate end of year exam so any solution shouldn't be too advanced.

$$I(a)=\int_0^{\pi/2}\frac{\sin(u+a\tan u)}{\sin u}du$$
by use this identitie
$$\sin(x+y)=\sin x \cos y + \cos x \sin y$$
so
$$I(a)=\int_0^{\pi/2}(\cos(a \tan u)+\frac{\sin(a\tan u)}{\tan u})du$$
let
$$u=\tan^{-1}v$$
then
$$I(a)=\int_0^{\infty}\frac{\cos(av)+\frac{\sin(av)}{v}}{(1+v^2)}dv$$
differentiate both side with respect to $a$
$$I'(a)=\int_0^{\infty}\frac{\cos(av)-v \sin (av)}{(1+v^2)}dv$$
both of 
$$\int_0^{\infty}\frac{\cos(av)}{(1+v^2)}dv ,,,,,,,,,\int_0^{\infty}\frac{v \sin (av)}{(1+v^2)}dv$$
can be shown here and here  in real method
thats lead to $$I'(a)=0$$
$$\Rightarrow I(a) \text{ is const}$$
$$I(a)=I(0)=\pi/2$$
