How to Solve a Trig Integral with Three Terms Using Integration by Parts?

[karush]

$\large {S6.7.1.17}$
$\tiny\text {trig Integration}$
$$\displaystyle
I= \int x \sec\left({x}\right)\tan\left({x}\right)\, dx \\$$
OK this has 3 terms so the inclination is IBP
If $u=\tan\left({x}\right)$ and $du=\sec^2 \left({x}\right) \, dx $
but don't see this fitting in
$\tiny\text{ Surf the Nations math study group}$

 

[Greg]

\(\displaystyle I=x\sec(x)-\int\sec(x)\,dx\)

 

[MarkFL]

$\large {S6.7.1.17}$
$\tiny\text {trig Integration}$
$$\displaystyle
I= \int x \sec\left({x}\right)\tan\left({x}\right)\, dx \\$$
OK this has 3 terms so the inclination is IBP
If $u=\tan\left({x}\right)$ and $du=\sec^2 \left({x}\right) \, dx $
but don't see this fitting in
$\tiny\text{ Surf the Nations math study group}$

LIATE tells us to try:

\(\displaystyle u=x\,\therefore\,du=dx\)

\(\displaystyle dv=\sec(x)\tan(x)\,dx\,\therefore\,v=\sec(x)\)

And this gives you the result Greg posted. :)

 

[karush]

$\large {S6.7.1.17}$
$\tiny\text {trig Integration}$
$$\displaystyle
I= \int x \sec\left({x}\right)\tan\left({x}\right)\, dx \\$$
IBP
$$\displaystyle
\begin{align}
{u} & = {x} & {dv}&= \sec\left({x}\right)\tan\left({x}\right)
\ d{x} & \\
{du}&=du& {v}&={\sec(x) }
\end{align} \\
I=x \sec(x) - \int \sec(x) dx \implies
x\sec\left(x\right)-\ln\left(\left|\tan\left(x\right)+\sec\left(x\right)\right|\right)+C
$$