Question

integration of root cotx

Answer 1

$\sqrt{Cot\ x} = \sqrt{ \frac{Cos\ x}{Sin\ x} } = \sqrt{ \frac{Cos\ x\ Cos\ x}{Sin\ x\ Cos\ x} } = \frac{Cos\ x}{2 \sqrt{Sin\ 2x}}\\ \\Let\ Sin\ 2x=t^2,\ \ \ 2\ Cos\ 2x\ dx=2t\ dt,\ \ \ (Cos^2x-sin^2x)dx=t\ dt\\ \\\frac{Cos\ x}{2\sqrt{Sin\ 2x}}dx=\frac{Cos\ x\ t\ dt}{2t(Cos\ x+sin\ x)(Cos\ x-Sin\ x)}=\\ \\=\frac{dt}{4(Cos\ x+Sin\ x)}+\frac{dt}{4(Cos\ x-Sin\ x)}\\ \\=\frac{dt}{4\sqrt{(Cos\ x+Sin\ x)^2}}+\frac{dt}{4\sqrt{(Cos\ x-Sin\ x)^2}}=$

$\frac{dt}{4\sqrt{Cos^2x+Sin^2x+Sin2x}}+\frac{dt}{4\sqrt{Cos^2x+Sin^2x-Sin2x}}=\\ \\\frac{dt}{4\sqrt{1+t^2}}+\frac{dt}{4\sqrt{1-t^2}}\\ \\ \int\limits^{}_{} {\sqrt{Cot\ x}} \, dx = \int\limits^{}_{} {\frac{1}{4\sqrt{1+t^2}}}} \, dt+\int\limits^{}_{} {\frac{1}{4\sqrt{1-t^2}}}} \, dt\\ \\\frac{1}{4}Ln|t+\sqrt{1+t^2}|+\frac{1}{4}Sin^{-1}t$

$=\frac{1}{4}Ln|t+\sqrt{1+t^2}|+\frac{1}{4}Sin^{-1}t\\ \\=\frac{1}{4}Ln|\sqrt{Sin\ 2x}+\sqrt{Cos^2x+Sin^2x+2Sinx\ Cos\ x}|+\frac{1}{4}Sin^{-1}\sqrt{Sin\ 2x}\\ \\=\frac{1}{4}Ln|\sqrt{Sin\ 2x}+Cosx+Sinx|+\frac{1}{4}Sin^{-1}\sqrt{Sin\ 2x}+C$