Evaluation of $ \int \tan x\cdot \sqrt{1+\sin x}dx$ Calculation of $\displaystyle \int \tan x\cdot \sqrt{1+\sin x}dx$
$\bf{My\; Try::}$ Let $\displaystyle (1+\sin x)= t^2\;,$ Then $\displaystyle \cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\sqrt{2-t^2}}dt$
So Integral is $\displaystyle  = \displaystyle 2\int \frac{t^2}{\sqrt{2-t^2}} \frac{(t^2-1)}{\sqrt{2-t^2}}dt = 2\int\frac{t^4-t^2}{2-t^2}dt $
Now How Can I solve after that
Help me
Thanks

I think you made some mistakes is the substitution,
$$t^2=\sin x +1$$ so 
$$2tdt=\cos x dx$$
now $$\frac{\tan x}{\cos x}= \frac{\sin x}{\cos^2 x}= \frac{\sin x}{1-\sin^2 x}= 
\frac{t^2-1}{1-(t^2-1)^2 }=\frac{t^2-1}{2t^2-t^4 }$$
So $$\int \tan x \sqrt{\sin x +1} dx= \int \frac{\tan x}{\cos x} \sqrt{\sin x +1} \cos x dx=\int \frac{t^2-1}{2t^2-t^4 }  2t^2dt=
\int 2\frac{t^2-1}{2-t^2 }  dt$$
To integrate 
$$\int \frac{t^2-1}{t^2-2}dt=\int 1+ \frac{1}{t^2-2}dt =
\int 1+ \frac{1}{2\sqrt{2}} \left( \frac{1}{t-\sqrt{2}} -\frac{1}{t-\sqrt{2}} \right)dt $$
Now we can integrate getting some logarithms.
