How do you find #int sec^2x/(1-sinx) #?

Question

2585 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-22T15:14:33

=#tanx +tan^3x/3+sec^3x/3+C#

#int(sec^2x)/(1-sinx)dx#
#=int(sec^2x)/(1-sinx)*(1+sinx)/(1+sinx)dx#

#=int(sec^2x)/(1-sin^2x)*(1+sinx)dx#

#=int(sec^2x/cos^2x)*(1+sinx)dx#

#=intsec^2x*(1/cos^2x+sinx/cos^2x)dx#
#=intsec^2x*sec^2xdx+intsec^2xsecxtanxdx#

#=intsec^2x*(1+tan^2x)dx+intsec^2xsecxtanxdx#
For first part let #u=tanx =>du=sec^2xdx#
For second part let #v = secx =>dv=secxtanxdx#

The whole integral becomes
=#int(1+u^2)du+intv^2dv#
=#u +u^3/3+v^3/3+C#

=#tanx +tan^3x/3+sec^3x/3+C#