How do you integrate #int xsec^2x# by parts from #[0, pi/4]#?

1 Answer
May 25, 2017

#1/4pi-1/2ln2#

Explanation:

First without the bounds:

#intxsec^2xcolor(white).dx#

Since integration by parts takes the form #intudv=uv-intvdu#, we need to assign values for #u# and #dv#. Let:

#{(u=x,=>,du=dx),(dv=sec^2xcolor(white).dx,=>,v=tanx):}#

So:

#intxsec^2xcolor(white).dx=xtanx-inttanxcolor(white).dx#

You may know this integral already, but if you're unsure how to find it, the process goes as such:

#-inttanxcolor(white).dx=int(-sinx)/cosxdx#

Let #t=cosx# so #dt=-sinxcolor(white).dx#:

#=int(dt)/t=lnabst=lnabscosx#

So:

#intxsec^2xcolor(white).dx=xtanx+lnabscosx#

Applying the bounds:

#int_0^(pi/4)xsec^2xcolor(white).dx=[xtanx+lnabscosx]_0^(pi/4)#

#=pi/4tan(pi/4)+lnabscos(pi/4)-( 0tan0+lnabscos0)#

#=pi/4(1)+lnabs(1/sqrt2)+(0+lnabs1)#

Using #1/sqrt2=2^(-1/2)# and #ln(a^b)=bln(a)#:

#=pi/4-1/2ln2#