Question

integrate the function 1/(1 + cotx).dx

Answer 1

you can see in it ok

Answer 2

I = ∫ 1/(1 + cotx) dx
= ∫ 1/(1 + cosx/sinx) dx
= ∫ sinx/(sinx + cosx) dx
= 1/2 ∫ 2sinx/(sinx + cosx) dx
= 1/2 ∫ {(sinx + cosx) + (sinx - cosx)}/(sinx + cosx) dx
= 1/2 ∫(sinx + cosx)/(sinx + cosx) dx + 1/2 ∫(sinx - cosx)/(sinx + cosx) dx
= 1/2 ∫dx + 1/2 ∫ (sinx - cosx)/(sinx + cosx) dx
= x/2 + C₁ + I₂

now, I₂ = 1/2 ∫ (sinx - cosx)/(sinx + cosx) dx
Let f(x) = sinx + cosx
differentiate both sides,
f'(x) = cosx - sinx = -(sinx - cosx)
so, I₂ = -1/2 ∫f'(x)/f(x) dx = -1/2lnf(x) + C₂
put f(x) = (sinx + cosx) in I₂ ,
hence, I₂ = -1/2 ln(sinx + cosx) + C₂

now, I = x/2 - 1/2 ln(sinx + cosx) + (C₁ + C₁)
= x/2 - ln(sinx + cosx) + C [ where C = C₁ +C₂]