What is the integral of 1/(1+cosx)?

2 Answers
Apr 2, 2018

I=-cotx+cscx+c

Explanation:

Note:
color(red)((1)intcosec^2thetad(theta)=-cottheta+c

color(red)((2)intcosecthetacotthetad(theta)=-cosectheta+c

We have,

I=int1/(1+cosx)dx

=int(1-cosx)/((1-cosx)(1+cosx))dx

=int(1-cosx)/(1-cos^2x)dx

=int(1-cosx)/sin^2xdx

=int(1/sin^2x-cosx/sin^2x)dx

=int(csc^2x-cscxcotx)dx....toApply (1) and(2)

=-cotx-(-cscx)+c

I=-cotx+cscx+c

Apr 2, 2018

One way to express the solution is:

int[1/(1+cosx)]dx=color(red)(cscx-cotx+C)

(See a solution process below)

Explanation:

int[1/(1+cosx)]dx

int[1/(1+cosx)*(1-cosx)/(1-cosx)]dx

int[(1-cosx)/color(red)(1-cos^2x)]dx

Using color(red)(1-cos^2x)=color(blue)(sin^2x)

int[(1-cosx)/color(blue)(sin^2x)]dx

int[1/sin^2x]dx-int[cosx/sin^2x]dx

Using cscx=1/sinx and cotx=cosx/sinx

int[csc^2x]dx-int[cscxcotx]dx

Integrating, we get:

[-cotx]+[cscx]+C

color(red)(cscx-cotx+C)