How do you evaluate the integral #∫ (1-sinx)/cosx dx#?

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Answer 1 · 2015-02-20T20:53:06

Well, this one is tough and I'll need to use a trick.

I multiply and divide by #1+sin(x)#!!!
So you get:

#int[(1-sin(x))(1+sin(x))]/[cos(x)(1+sin(x))]dx=#

#=int[1-sin^2(x)]/[cos(x)(1+sin(x))]dx=#

#=int[cos^2(x)]/[cos(x)(1+sin(x))]dx=#

#=int[cos(x)]/[1+sin(x)]dx=#

but #d[sin(x)]=cos(x)dx#

and:
#=int[1]/[1+sin(x)]d[sin(x)]=# and finally:

#=ln|1+sin(x)|+c#

Hope it helps