How do you integrate #int x/ cos^2(x)dx#?

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Answer 1 · 2016-09-10T20:28:17

#xtan(x)+ln(abscos(x))+C#

We have:

#I=x/cos^2(x)dx=intxsec^2(x)dx#

We will use integration by parts, which takes the form #intudv=uv-intvdu#. For the given integral #intxsec^2(x)dx#, let:

#{(u=x,(du)/dx=1,du=dx),(dv=sec^2(x)dx,intdv=intsec^2(x)dx,v=tan(x)):}#

Thus:

#I=xtan(x)-inttan(x)dx#

#I=xtan(x)+int(-sin(x))/cos(x)dx#

Letting #y=cos(x)# such that #dy=-sin(x)dx#:

#I=xtan(x)+int(dy)/y#

#I=xtan(x)+ln(absy)#

#I=xtan(x)+ln(abscos(x))+C#