How do you integrate #int x/ cos^2(x)dx#?
1 Answer
Sep 10, 2016
Explanation:
We have:
#I=x/cos^2(x)dx=intxsec^2(x)dx#
We will use integration by parts, which takes the form
#{(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
#I=xtan(x)+int(dy)/y#
#I=xtan(x)+ln(absy)#
#I=xtan(x)+ln(abscos(x))+C#