How to integrate 1/(1+cosx) ?

May 6, 2018

$I = \csc x - \cot x + c$

Explanation:

We know that,

color(violet)((1)sin^2theta+cos^2theta=1

color(blue)((2)1/sintheta=csctheta and costheta/sintheta=cottheta

color(red)((3)int(csc^2x)dx=-cotx+c

color(green)((4)int(cscxcotx)dx=-cscx+c
Here,

$I = \int \frac{1}{1 + \cos x} \mathrm{dx}$

$= \int \frac{1}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x} \mathrm{dx}$

$= \int \frac{1 - \cos x}{1 - {\cos}^{2} x} \mathrm{dx}$

=int(1-cosx)/sin^2xdx...tocolor(violet)(Apply(1)

$= \int \left[\frac{1}{\sin} ^ 2 x - \cos \frac{x}{\sin} ^ 2 x\right] \mathrm{dx}$

=int[csc^2x-cscxcotx]dx...tocolor(blue)(Apply(2)

=-cotx-(-cscx)+c...toApplycolor(red)((3))andcolor(green)((4)

$I = \csc x - \cot x + c$

May 6, 2018

$I = \tan \left(\frac{x}{2}\right) + c$

Explanation:

$I = \int \frac{1}{1 + \cos x} \mathrm{dx}$

Let, $\tan \left(\frac{x}{2}\right) = t \implies {\sec}^{2} \left(\frac{x}{2}\right) \times \frac{1}{2} \mathrm{dx} = \mathrm{dt} \implies \mathrm{dx} = \frac{2 \mathrm{dt}}{1 + {\tan}^{2} t}$

=>dx=(2dt)/(1+t^2)andcosx=(1-tan^2(x/2))/(1+tan^2(x/2))=(1- t^2)/(1+t^2)

So,

$I = \int \frac{1}{1 + \frac{1 - {t}^{2}}{1 + {t}^{2}}} \times \frac{2}{1 + {t}^{2}} \mathrm{dt}$

$= \int \frac{2}{1 + {t}^{2} + 1 - {t}^{2}} \mathrm{dt}$

$= \int 1 \mathrm{dt} = t + c , w h e r e , t = \tan \left(\frac{x}{2}\right)$

$I = \tan \left(\frac{x}{2}\right) + c$

Note : Both the answer are same ,but in different form.

$\csc x - \cot x = \frac{1}{\sin} x - \cos \frac{x}{\sin} x$

$= \frac{1 - \cos x}{\sin} x$

=(2sin^2(x/2))/(2sin(x/2)cos(x/2)

$= \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right)$

$= \tan \left(\frac{x}{2}\right)$