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
 A: Performing a polynomial long division followed by a partial fraction decomposition results in
$$2\int\frac{t^4-t^2}{2-t^2}dt=-2\int(t^2+1)dt+\sqrt{2}\int\left(\frac{1}{\sqrt{2}-t}+\frac{1}{\sqrt{2}+t}\right)dt$$
I suppose you can take it from there.
A: 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.
A: $$ 2\int\frac{t^4-t^2}{2-t^2}dt = 2\int\frac{\require{cancel}\cancel{(t^2 - 2)}(t^2 + 1)+2}{-(\cancel{t^2 - 2})},dt = \underbrace{-2\int (t^2 + 1)\,dt}_{\text{a cinch}} - \underbrace{2\int \frac {2}{t^2 - 2}\,dt}_{\text{partial fractions}}$$
Alternatively, for the second integral, we can express it as $$+ 2\cdot\frac{2}{1-t^2}$$ and use $t = \sin \theta \,d\theta \implies dt = \cos \theta \,d\theta $ to get $$4\int \sec \theta d\,\theta$$
A: Hint: $\frac{t^4-t^2}{2-t^2} \equiv -t^2-\frac{2}{t^2-2}-1,$ which is easy to integrate.


 You should find, after a suitable hyperbolic substitution, that $\int \frac{2}{t^2-2}dt= \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2}-t}{\sqrt{2}+t} \right|$.

