Integrate 1+Tanx tan(x+theta) dx

1.	Integrate 1+Tanx tan(x+theta) dx
Answer» ∫ \(\frac{1+tan\,x\,.\,tan(x+\theta)}{[tan(x+\theta)-tan\,x]}\) (tan(x+θ) - tan x)dx = ∫ \(\frac1{tan\,\theta}\) [tan(x+θ) - tan x] . dx = \(\frac1{tan\,\theta}\) ∫ tan(x+θ)dx - ∫ tan x . dx = \(\frac1{tan\,\theta}\) [-log cos(x+θ)1] + [log \|cos x\|] + c = \(\frac1{tan\,\theta}\) log \|\(\frac{cos\,x}{cos(x+\theta)}\)\| + c = cot θ log \|\(\frac{cos\,x}{cos(x+\theta)}\)\| + c.

Answer»

∫ \(\frac{1+tan\,x\,.\,tan(x+\theta)}{[tan(x+\theta)-tan\,x]}\) (tan(x+θ) - tan x)dx

= ∫ \(\frac1{tan\,\theta}\) [tan(x+θ) - tan x] . dx

= \(\frac1{tan\,\theta}\) ∫ tan(x+θ)dx - ∫ tan x . dx

= \(\frac1{tan\,\theta}\) [-log cos(x+θ)1] + [log |cos x|] + c

= \(\frac1{tan\,\theta}\) log |\(\frac{cos\,x}{cos(x+\theta)}\)| + c

= cot θ log |\(\frac{cos\,x}{cos(x+\theta)}\)| + c.

Discussion