e-x-1-sin-x-1-cos-x-dx-

Question Number 59344 by rahul 19 last updated on 08/May/19

\int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x = ?

Answered by MJS last updated on 08/May/19

the trick is this :

\frac{1 - \sin x}{1 - \cos x} = \frac{1}{1 - \cos x} - \frac{\sin x}{1 - \cos x} = \frac{1}{{2 \sin}^{2} \frac{x}{2}} - \frac{1}{\tan \frac{x}{2}} =

= \frac{1}{2} \csc^{2} \frac{x}{2} - \cot \frac{x}{2}

so we have

\int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x = \frac{1}{2} \int e^{x} \csc^{2} \frac{x}{2} d x - \int e^{x} \cot \frac{x}{2} d x

now solve the 2^{nd} one by parts

\int f^{'} g = f g - \int {f g}^{'}

f^{'} = f = e^{x}

g = \cot \frac{x}{2} \Rightarrow g^{'} = - \frac{\csc^{2} x}{2}

\int e^{x} \cot \frac{x}{2} d x = e^{x} \cot \frac{x}{2} + \frac{1}{2} \int e^{x} \csc^{2} \frac{x}{2} d x

\Rightarrow

\int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x = - e^{x} \cot \frac{x}{2} + C

Commented by rahul 19 last updated on 08/May/19

t h a n k y o u s i r .

Commented by MJS last updated on 08/May/19

you're welcome; please check my answers before using them because I'm a typo-lover ��

Commented by rahul 19 last updated on 08/May/19

well, I already have answers to almost all the Q., I post here! All I want is "trick" .��

Commented by malwaan last updated on 09/May/19

f a n t a s t i c s i r!

w h a t i s t y p o - l o v e r ?

Answered by tanmay last updated on 08/May/19

\int e^{x} (\frac{1 - 2 s i n \frac{x}{2} c o s \frac{x}{2}}{2 {s i n}^{2} \frac{x}{2}}) d x

\int (e^{x} \times \frac{{c o s e c}^{2} \frac{x}{2}}{2} - e^{x} \times c o t \frac{x}{2}) d x

(- 1) \int e^{x} c o t \frac{x}{2} - e^{x} \times \frac{{c o s e c}^{2} \frac{x}{2}}{2} d x

(- 1) \int [c o t \frac{x}{2} \times \frac{{d e}^{x}}{d x} + e^{x} \times \frac{d}{d x} (c o t \frac{x}{2})] d x

(- 1) \int \frac{d}{d x} (e^{x} c o t \frac{x}{2}) d x

- e^{x} c o t \frac{x}{2} + c

Commented by rahul 19 last updated on 09/May/19

thank you sir!