Question

integrate under root 10 x + under root cot X

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