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

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Answer 1 · 2018-04-02T16:49:04

#I=-cotx+cscx+c#

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#

Answer 2 · 2018-04-02T16:50:30

One way to express the solution is:

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

(See a solution process below)

#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)#