Can You Outsmart This Math Challenge? | An Integration Problem

[@Quest3669]

√2sin^-1(sinx-cosx)+c

[@gregevgeni1864]

The key, t = sin x - cos x .
√tan x + √cot x = √((sin x)/cos x)) + √((cos x)/(sin x)) ={(√sin x)^2) + (√cos x)^2}/√(sin x)(cos x) =
(sin x + cos x)/√(sin x)(cos x) (*)
So the original integral written
Int {(sin x + cos x)/√(sin x)(cos x)} dx = K, due to (*) .
Put t = sin x - cos x .(**)
Then dt = (sin x + cos x) dx,
t^2 = (sin x - cos x)^2 =
= 1 - 2 (sin x)(cos x) =>
(sin x)(cos x) = (1 -t^2)/2 .
Hence K = Int( dt /(√(1 -t^2)/2) =
= Int ( dt / (√(1 - t^2)/√2) =
= √2 Int ( dt /√(1 -t^2)=
√2 arc sin(sin t) + c =
= √2 arc sin( sin x - cos x) + c.

[@RashmiRay-c1y]

The integral can be written as \int dx [(sin x+ cos x)/√(sin x cos x)] = √2 \int dx [(sin x+ cos x)/√(sin 2x))] =2 \int dx [(sin x+ cos x)/√{1-(1-sin2 x)}] = 2 \int dx [(sin x+ cos x)/√{1-(sinx -cos x)^2}]. Let t-sin x - cos x . Then, dt = dx(sin x + cos x) and so I = √2 \int dt /√(i-t^2) = √2 arcsin t + C = √2 arcsin (sin x -cos x) + C. Thus, I = √2 arcsin (sin x -cos x) + C.

[@michaeldoerr5810]

This is honestly BETTER than the comments made. And I shall use that as a challenge. Juat to redeem myself of any doubts on doing calculus. And I think that I have understood it!!! I hope that thia means that I up for the challenge!!!

[@RyanLewis-Johnson-wq6xs]

integral(sqrt(tan(x)) + sqrt(cot(x))) dx = Sqrt[2] tan^(-1)(t-1/t)/Sqrt[2]+C

[@michaeldoerr5810]

CORRECT!!! That was what I got!!!