Question

integration of root cotx

Answer 1

∫√cot x dx =½∫[(√cot x+√tan x)+(√cot x−√tan x)] dx =½∫[(cos x+sin x)/√(cos x sin x)] dx +½∫[(cos x−sin x)/√(cos x sin x)] dx =(1/√2)∫[(cos x+sin x)/√sin 2x] dx +(1/√2)∫[(cos x−sin x)/√sin 2x] dx …..(i) [since 2 cos x sin x=sin 2x] Suppose sin 2x= z² Therefore, cos 2x dx= z dz or, (cos² x−sin² x) dx= z dz or, (cos x+sin x)(cos x−sin x) dx= z dz Now (cos x+sin x) dx =z dz/(cos x−sin x) =z dz/√(cos² x+sin² x−2 cos x sin x) =z dz/√(1−sin 2x) =z dz/√(1−z²) …..(ii) Similarly, (cos x−sin x) dx =z dz/√(1+z²) …..(iii) From (i), (ii) and (iii), ∫√cot x dx =(1/√2)∫z dz/z√(1−z²)+(1/√2)∫z dz/z√(1+z²) =(1/√2)∫dz/√(1−z²)+(1/√2)∫dz/√(1+z²) =(1/√2)[∫dz/√(1−z²)+∫dz/√(1+z²)] =(1/√2)[sin-1 z+ln |z+√(1+z²)|]+c =(1/√2)[sin-1 √sin 2x+ln |√sin 2x+√(1+sin 2x)|]+c =(1/√2)[sin-1 √sin 2x+ln |√sin 2x+√(cos² x+sin² x +sin 2x)|]+c =(1/√2)[sin-1 √sin 2x+ln |√sin 2x+cos x+sin x|]+c

Answer 2

Integral of  √cotx ,  or √tanx  or   x²/(1+x⁴)   or of  √cotx - √tanx  or  √cotx + √tanx  are all related and similar....

Let cotx = t²   =>   -(1+t⁴) dx=2t dt 

Let      y = t - 1/t     so    dy = (1 + 1/t²) dt
Let      z = t + 1/t      so    dz = (1 - 1/t²)  dt
Also   1/(z² - 2) = (1/2√2) * [ 1/(z-√2) - 1/(z+√2)]   using partial fractions.

$I = \int {\sqrt{cotx}} \ dx= - \int {\frac{2t^2}{1+t^4}} \ dt\\\\=- \int {\frac{t^2+1}{1+t^4}} \ dt - \int {\frac{t^2-1}{1+t^4}} \ dt\\\\=- \int {\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-2+2}} dt - \int { \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}+2-2}} \ dt\\\\=- \int {\frac{dy}{y^2+2}} - \int {\frac{dz}{z^2-2}} \\\\=-\frac{1}{\sqrt2} tan^{-1}[\frac{y}{\sqrt2}]-\frac{1}{2\sqrt2}Ln | \frac{z-\sqrt2}{z+\sqrt2} | +c$

$I=-\frac{1}{\sqrt2} tan^{-1}[\frac{cotx-1}{\sqrt{2cotx}}]-\frac{1}{2\sqrt2}Ln | \frac{cotx+1-\sqrt{2cotx}}{cotx+1+\sqrt{2cotx}} | +c$