Question is of integration .

Question

Integration of 1/ 1+tanx.

Answer 1

Hello Shubham

There are many different ways of answering this question, some easy to understand and others more difficult to understand. I will provide you with an easy and understandable way of solving this problem.

∫{1/[1 + tan(x)]}dx = ∫{1/[cos(x)/cos(x) + sin(x)/cos(x)]}dx = ∫{1/([sin(x) + cos(x)]/cos(x))}dx =

∫{cos(x)/[sin(x) + cos(x)]}dx = (1/2)∫{2cos(x)/[sin(x) + cos(x)]}dx = (1/2)∫({2cos(x)/[sin(x) + cos(x)]} - 1 + 1)dx =

(1/2)∫({[cos(x) + cos(x)]/[sin(x) + cos(x)] - [sin(x) + cos(x)]/[sin(x) + cos(x)] } + 1)dx =

(1/2)∫({[cos(x) + cos(x) - sin(x) - cos(x)]/[sin(x) + cos(x)] } + 1)dx = (1/2)∫{[cos(x) - sin(x)]/[sin(x) + cos(x)]}dx + (1/2)∫dx. Let u = sin(x) + cos(x). Therefore, du = [cos(x) - sin(x)]dx = (1/2)∫(1/u)du + (1/2)∫dx =

(1/2)ln|u| + (1/2)x + c = (1/2)ln|sin(x) + cos(x)| + (1/2)x + c.

If you have further questions, please feel free to contact me.

Sincerely

Michael I.