# How do you integrate int xsec^-1x by integration by parts method?

Mar 3, 2017

This is a problem that will necessitate integration by parts. We let $u = {\sec}^{-} 1 x$ and $\mathrm{dv} = x$. We can easily see that $v = \frac{1}{2} {x}^{2}$, but we will have to work to find the derivative of $u$.

$u = {\sec}^{-} 1 x$

$\sec u = x$

$\sec u \tan u \frac{\mathrm{du}}{\mathrm{dx}} = 1$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{\sec u \tan u}$

Because $x = \sec u$, this means that the side adjacent $u$, if we were to draw an imaginary triangle, would measure $1$ and the hypotenuse would measure $x$.

This means the side opposite would measure $\sqrt{{x}^{2} - 1}$, and so $\tan u = \frac{\sqrt{{x}^{2} - 1}}{1} = \sqrt{{x}^{2} - 1}$.

(du)/dx = 1/(xsqrt(x^2 - 1)

$\mathrm{du} = \frac{1}{x \sqrt{{x}^{2} - 1}} \mathrm{dx}$

We now have, by the integration by parts formula,

$\int \left(u \mathrm{dv}\right) = u v - \int \left(v \mathrm{du}\right)$

$\int x {\sec}^{-} 1 x \mathrm{dx} = {\sec}^{-} 1 x \left(\frac{1}{2} {x}^{2}\right) - \int \frac{1}{2} {x}^{2} \frac{1}{x \sqrt{{x}^{2} - 1}} \mathrm{dx}$

$\int x {\sec}^{-} 1 x \mathrm{dx} = {\sec}^{-} 1 x \frac{1}{2} {x}^{2} - \frac{1}{2} \int \frac{x}{\sqrt{{x}^{2} - 1}} \mathrm{dx}$

We could use trig substitution to integration $\frac{x}{\sqrt{{x}^{2} - 1}}$. However, in this case, a u-substitution would do the trick easily.

Let $u = {x}^{2} - 1$. Then $\mathrm{du} = 2 x \mathrm{dx} \to \mathrm{dx} = \frac{\mathrm{du}}{2 x}$.

$\int x {\sec}^{-} 1 x \mathrm{dx} = {\sec}^{-} 1 x \frac{1}{2} {x}^{2} - \frac{1}{2} \int \frac{x}{\sqrt{u}} \cdot \frac{\mathrm{du}}{2 x}$

$\int x {\sec}^{-} 1 x \mathrm{dx} = {\sec}^{-} 1 x \frac{1}{2} {x}^{2} - \frac{1}{4} \int {u}^{- \frac{1}{2}} \mathrm{du}$

$\int x {\sec}^{-} 1 x \mathrm{dx} = \frac{1}{2} {x}^{2} {\sec}^{-} 1 x - \frac{1}{4} \left(2 {u}^{\frac{1}{2}}\right) + C$

$\int x {\sec}^{-} 1 x \mathrm{dx} = \frac{1}{2} {x}^{2} {\sec}^{-} 1 x - \frac{1}{2} {u}^{\frac{1}{2}} + C$

$\int x {\sec}^{-} 1 x \mathrm{dx} = \frac{1}{2} {x}^{2} {\sec}^{-} 1 x - \frac{1}{2} {\left({x}^{2} - 1\right)}^{\frac{1}{2}} + C$

Hopefully this helps!