sec-d-1-sec-

Question Number 21679 by Arnab Maiti last updated on 30/Sep/17

\int \frac{\sec θ d θ}{1 - \sec θ}

Answered by alex041103 last updated on 30/Sep/17

F i r s t w e m a k e t h e f o l l o w i n g t r a n s f o r m a t i o n s :

\int \frac{\sec θ d θ}{1 - \sec θ} = \int \frac{\frac{1}{c o s θ}}{1 - \frac{1}{c o s θ}} d θ =

= \int \frac{\frac{1}{c o s θ}}{1 - \frac{1}{c o s θ}} \frac{c o s θ}{c o s θ} d θ =

= \int \frac{1}{c o s θ - 1} d θ

A l s o b y t h e d o u b l e a n g l e i n d e t i t y

w e h a v e c o s θ = 1 - 2 {s i n}^{2} (\frac{θ}{2})

\Rightarrow \int \frac{1}{c o s θ - 1} d θ = - \int \frac{1}{{s i n}^{2} (\frac{θ}{2})} \frac{d θ}{2}

L e t x = \frac{θ}{2} \Rightarrow d x = \frac{d θ}{2}

\int \frac{1}{c o s θ - 1} d θ = \int - {c s c}^{2} x d x = I

W e k n o w t h a t \frac{d}{d x} (c t g x) = - {c s c}^{2} x

\Rightarrow \int \frac{s e c θ}{1 - s e c θ} d θ = c t g (\frac{θ}{2}) + C