Integration by Parts problem

**bijoustrollette** · March 23rd 2010, 12:18 PM

Hi guys, I have to teach a lesson on this tomorrow (trainee teacher, gulp!) and can't figure out this problem.

Integrate:

x^2sec^2xtanx.dx

I can't seem to figure out if there's a way to manipulate this to make integration by parts easier. Any ideas? Would making it x^2sec.secxtanx help at all given that secxtanx integrates to sec x?

Thanks in advance for any help.

**integral** · March 23rd 2010, 12:58 PM

Hint: $Sec^2(x)=tan^2(x)+1$

**ANDS!** · March 23rd 2010, 01:06 PM

Ouch. This - is a long one. You will need to use integration by parts twice - it is painless though - and a substitution in one of your parts:

$\int x^{2}sec^{2}(x)tan(x)dx$

$u = x^{2}, du=2x$

$\int dv = \int sec(x)sec(x)tan(x)dx$

Using substitution:

$w=sec(x), dw=sec(x)tan(x)dx$

$v = \frac{sec^{2}(x)}{2}$

Setting up the by parts:

$\int x^{2}sec^{2}(x)tan(x)dx = \frac{x^{2}sec^{2}(x)}{2} -\int xsec^{2}(x)dx$

Taking integration by parts on the last integral:

$u = x, du = 1dx$

$dv = sec^{2}(x), v = tan(x)$

Then (I'm only going to do the last part):

$\int xsec^{2}(x)dx = xtan(x) - \int tan(x)dx \Rightarrow$

$\int xsec^{2}(x)dx = xtan(x) +ln |cos(x)|$

All together now:

$\int x^{2}sec^{2}(x)tan(x)dx = \frac{x^{2}sec^{2}(x)}{2} - xtan(x) - ln |cos(x)|$

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