How do you integrate #int xsec^2x# by parts from #[0, pi/4]#?
1 Answer
Explanation:
First without the bounds:
#intxsec^2xcolor(white).dx#
Since integration by parts takes the form
#{(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
#=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
#=pi/4-1/2ln2#