How to integrate #1/(1+cosx)# ?

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Answer 1 · 2018-05-06T03:02:48

#I=cscx-cotx+c#

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=int1/(1+cosx)dx#

#=int1/(1+cosx)xx(1-cosx)/(1-cosx)dx#

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

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

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

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

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

#I=cscx-cotx+c#

Answer 2 · 2018-05-06T12:31:30

#I=tan(x/2)+c#

#I=int1/(1+cosx)dx#

Let, #tan(x/2)=t=>sec^2(x/2)xx1/2dx=dt=>dx=(2dt)/(1+tan^2t)#

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

So,

#I=int1/(1+(1-t^2)/(1+t^2))xx2/(1+t^2)dt#

#=int2/(1+t^2+1-t^2)dt#

#=int1dt=t+c,where,t=tan(x/2)#

#I=tan(x/2)+c#

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

#cscx-cotx=1/sinx-cosx/sinx#

#=(1-cosx)/sinx#

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

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

#=tan(x/2)#