Question

integrate the function √tanx/sinxcosx.dx

Answer 1

you can see your answer

Answer 2

$\bf{\int{\frac{\sqrt{tanx}}{sinx.cosx}}\,dx}$
multiplying cosx both numerator and denominator
$\bf{\int{\frac{\sqrt{tanx}.cosx}{sinx.cos^2x}}\,dx}\\=\bf{\int{\frac{\sqrt{tanx}.sec^2x}{\frac{sinx}{cosx}}}\,dx}\\=\bf{\int{\frac{\sqrt{tanx}.sec^2x}{tanx}}\,dx}\\=\bf{\int{\frac{sec^2x}{\sqrt{tanx}}}\,dx}$

Let tanx = P.
differentiate with respect to x ,
sec²x.dx = dP
so, $\bf{\int{\frac{sec^2x}{\sqrt{tanx}}}\,dx}=\int{\frac{dt}{\sqrt{t}}}$
= $2\sqrt{t}+C$
now, put t = tanx
so, I = $\bf{2\sqrt{tanx}+C}$