How do you integrate #(1-cosx)/(1+cosx)#?

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Answer 1 · 2015-04-17T18:18:03

First just do simple math :

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

Now we can factorize the numerator :

#=>int(-(1+cos(x))+2)/(1+cos(x))dx#

#=>int(-(1+cos(x)))/(1+cos(x))dx+2int1/(1+cos(x))dx#

#=>int -1dx+2int1/(1+cos(x))dx#

Remember that #cos^2(x) = 1/2(1+cos(2x))#

(From #cos(a)cos(b)# formula with #a = b#)

#=>2cos^2(x) = (1+cos(2x))#

#=>2cos^2(1/2x) = 1+cos(x)#

#=>int -1dx+2int1/2*1/cos^2(1/2x)dx#

#1/2*1/cos^2(1/2x)dx# is exactly the derivate of #tan(1/2x)#

#=>[2tan(1/2x) - x] + C#