How do you evaluate the integral #int tan^2x/secx#?
1 Answer
Feb 2, 2017
Explanation:
We see that:
#inttan^2x/secxdx=int(sin^2x/cos^2x)/(1/cosx)dx=intsin^2x/cosxdx=int(1-cos^2x)/cosxdx#
Splitting up the integral:
#=intsecxdx-intcosxdx#
Both of these are standard integrals:
#=ln(abs(secx+tanx))-sinx+C#
