Asked by bhavesh

I=∫(√tanx+√cotx)dx

let √tanx=t

dt = 1/(2√tanx)sec²x = {1+t⁴}/2t.dx

dx = 2t/(1+t⁴)dt

I=∫(√tanx+√cotx)dx

subst from above,

I=∫(t+1/t)2t/(1+t⁴)dt

I=∫2(t²+1)/(t⁴+1)dt

now, divide numerator and denominator by t²

I=∫2(1/t²+1)/(t²+1/t²)dt

I=∫2(1/t²+1)/{(t-1/t)²)+2}dt

Now, t-1/t = m

differentiating, (1+1/t)dt = dm

Subst in equation above...

I = ∫2/{m²+2}dm

I =2/√2. tan^-1(m/√2)

I = √2 tan^-1(m/√2)

Where m=√tan x-√cot x

good work dude :)