Question

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

Answer 1

`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`