Question

integrate (sqrt tanx+ sqrt cotx) dx

Answer 1

Integration of sqrt(tan x) + sqrt(cot x) dx

$f(x)=\sqrt{tan x}+\sqrt{cot x}=\frac{sin x+cos x}{\sqrt{sin x\ cos x}}\\\\=\sqrt{2}*\frac{sinx+cosx}{\sqrt{sin2x}}\\\\Let\ t^2=sin2x\ ;; \ \ 2t\ dt=2cos2x\ dx=2\sqrt{1-t^4}dx\\\\I=\sqrt{2}\int {f(x)} \, dx=\sqrt{2}\int {\frac{\sqrt{1+t^2}}{t}\frac{t}{\sqrt{1-t^4}}} \, dt=\sqrt{2}\int {\frac{1}{\sqrt{1-t^2}}} \, dt\\\\=\sqrt2\ Sin^{-1}t+K=\sqrt2\ Sin^{-1}(\sqrt{sin 2x})+K\\\\=\sqrt{2}\ Tan^{-1}(\frac{cosx-sinx}{\sqrt{sin2x}})+K\\\\=\sqrt2\ Tan^{-1} (\frac{\sqrt{cotx}-\sqrt{tanx}}{\sqrt2})+K$


We can find Integral of  √cot x    or   of √tan x    or  of   √cot x - √tan x
In the same manner as above.

Answer 2

Answer:(sqrt2)arcsin(sinx-cosx)+C

Step-by-step explanation: