How do you evaluate the integral #int xsec^2x#?

1 Answer
mason m
Jan 14, 2017

#intxsec^2(x)dx=xtan(x)+ln(abscos(x))+C#

Explanation:

This is a prime candidate for integration by parts, which takes the form #intudv=uv-intvdu#.

For the given integral #intxsec^2(x)dx#, we want to choose a value of #u# that gets simpler when we differentiate it and a value of #dv# that is easily integrated.

So, let:

#{(u=x" "=>" "du=dx),(dv=sec^2(x)dx" "=>" "v=tan(x)):}#

We then have:

#intxsec^2(x)dx=uv-intvdu#

#color(white)(intxsec^2(x)dx)=xtan(x)-inttan(x)dx#

You may have the integral of #tan(x)# memorized. If not, it's easy to find:

#color(white)(intxsec^2(x)dx)=xtan(x)-intsin(x)/cos(x)dx#

Let #t=cos(x)#, implying that #dt=-sin(x)dx#:

#color(white)(intxsec^2(x)dx)=xtan(x)+int(-sin(x))/cos(x)dx#

#color(white)(intxsec^2(x)dx)=xtan(x)+int1/tdt#

This is a common integral:

#color(white)(intxsec^2(x)dx)=xtan(x)+ln(abst)+C#

Working back from #t=cos(x)#:

#color(white)(intxsec^2(x)dx)=xtan(x)+ln(abscos(x))+C#

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