If $\alpha$= $\int_{\pi/6}^{\pi/3}$ ($sint+cost$)/$\sqrt {sin2t}$ $dt$ then find the value of $f(\alpha)$= ($2sin({\alpha/2})+1)^2$.
My approach:
$\alpha$= $\int_{\pi/6}^{\pi/3}$ $\sqrt {(1+sin2t)/sin2t}$ $dt$ =$\int_{\pi/6}^{\pi/3}$ $\sqrt {(1+cosec2t)} dt$ = $\int_{\pi/6}^{\pi/3}$ $(\frac {cot2t}{\sqrt {cosec2t-1}})dt$.
We put $cosec2t-1$=$z^2$ $\Rightarrow$ $-cosec2tcot2tdt$= $zdz$ $\Rightarrow$ $cot2tdt$= $(\frac {-zdz}{{1+z^2}})$.
So now $\alpha$= $\int$ $\frac {-dz}{(1+z^2)}$= $-\arctan $ $z$ =$-\arctan (\sqrt{cosec2t-1})$ .
After putting the limits, I am getting $\alpha$=$0$. So $f(\alpha)$=$1$. But the answer is given $3$. Where have I done wrong? Please help.
Hint:
As $\int(\sin t+\cos t)dt=\sin t -\cos t+K$
choose $\sin t -\cos t=u\implies u^2=?$