Question

integration of tanx^1÷2

Answer 1

I$I = \int\limits {\sqrt{tanx}} \, dx$
we directly can't  find out this integration. by using some step we can do it .
step 1:-  find integration of 
$\int(\sqrt{tanx} +\sqrt{cotx}).dx$
$\int\frac{(sinx+cosx)}{\sqrt{sinx.cosx}}.dx\\=\sqrt{2}\int\frac{(sinx+cosx)}{\sqrt{sin2x}}$
             let z = $\int(sinx+cosx)dx$
                    = -cosx +sinx 
dz = (sinx+cosx).dx 
                now use this results in above,
then integration convert in
            I = $\sqrt{2}\int\frac{dz}{\sqrt{1-z^2}}$
              =$\sqrt{2}sin^{-1}z\\=\sqrt{2}sin^{-1}(sinx-cosx)$
step2:- similarly find out integration of 
    I = $\int(\sqrt{tanx}-\sqrt{cotx}).dx$
      = $\sqrt{2}ln(-sinx-cosx)+\sqrt{(-sinx-cosx)^2-1})$

step 3:- now add both inegrations .

i.e.  I =$\int(\sqrt{tanx}+\sqrt{cotx})dx+\int(\sqrt{tanx}-\sqrt{cotx})dx\\=2\int{\sqrt{tanx}.dx$

so, finally inegration we get ,
I = $\int\sqrt{tanx}dx$
=$=\frac{1}{\sqrt{2}} [sin^{-1}(sinx-cosx)+ln(-sinx-cosx+\sqrt{(-sinx-cosx)^2-1})]$