# How do you integrate int x arcsec x  using integration by parts?

May 12, 2018

$= {x}^{2} / 2 \setminus a r c \sec x - \frac{1}{2} \sqrt{{x}^{2} - 1} + C$

#### Explanation:

Clearly, you must need to know ${\left(a r c \sec \left(x\right)\right)}^{'}$ to do this by IBP.

So let $y = a r c \sec \left(x\right)$:

• $\sec y = x$

• ${\left(\sec y = x\right)}^{'} \implies \sec y \setminus \tan y \setminus y ' = 1$

• $\implies y ' = \frac{1}{\sec y \setminus \tan y} = \frac{1}{x \sqrt{{x}^{2} - 1}} \textcolor{red}{= {\left(a r c \sec \left(x\right)\right)}^{'}}$

Next is IBP:

$\int f ' \left(x\right) g \left(x\right) \setminus \mathrm{dx} = f \left(x\right) \setminus g \left(x\right) - \int f \left(x\right) g ' \left(x\right) \setminus \mathrm{dx}$

$\implies \int \mathrm{dx} \setminus x \setminus a r c \sec x = \int \mathrm{dx} \setminus {\left({x}^{2} / 2\right)}^{'} \setminus a r c \sec x$

$= {x}^{2} / 2 \setminus a r c \sec x - \int \mathrm{dx} \setminus {x}^{2} / 2 \setminus \frac{1}{x \sqrt{{x}^{2} - 1}}$

$= {x}^{2} / 2 \setminus a r c \sec x - \frac{1}{2} \int \mathrm{dx} \setminus \frac{x}{\sqrt{{x}^{2} - 1}} = \triangle$

Given that: ${\left(\sqrt{{x}^{2} - 1}\right)}^{'} = \frac{1}{2} \cdot \frac{1}{\sqrt{{x}^{2} - 1}} \cdot 2 x = \frac{x}{\sqrt{{x}^{2} - 1}}$

Then:

$\triangle = {x}^{2} / 2 \setminus a r c \sec x - \frac{1}{2} \int \mathrm{dx} \setminus {\left(\sqrt{{x}^{2} - 1}\right)}^{'}$

$= {x}^{2} / 2 \setminus a r c \sec x - \frac{1}{2} \sqrt{{x}^{2} - 1} + C$

May 12, 2018

$\int \setminus x \setminus \text{arcsec" \ x \ dx = 1/2x^2 \ "arcsec} \setminus x - \frac{1}{2} \sqrt{{x}^{2} - 1} + c$

#### Explanation:

We seek:

$I = \int \setminus x \setminus \text{arcsec} \setminus x \setminus \mathrm{dx}$

We can then apply Integration By Parts:

Let  { (u,="arcsec" \ x, => (du)/dx,=1/(xsqrt(x^2-1))), ((dv)/dx,=x, => v,=1/2x^2 ) :}

Then plugging into the IBP formula:

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

We have:

$\int \setminus \left(\text{arcsec" \ x)(x) \ dx = ("arcsec} \setminus x\right) \left(\frac{1}{2} {x}^{2}\right) - \int \setminus \left(\frac{1}{2} {x}^{2}\right) \left(\frac{1}{x \sqrt{{x}^{2} - 1}}\right) \setminus \mathrm{dx}$

$I = \frac{1}{2} {x}^{2} \setminus \text{arcsec} \setminus x - \frac{1}{2} \setminus \int \setminus \frac{x}{\sqrt{{x}^{2} - 1}} \setminus \mathrm{dx}$

For the integral IBP has introduced, we can perform a substitution, Let:

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

And if we substitute this into the integral we get:

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

And restoration of the substitution gives us:

$\int \setminus \frac{x}{\sqrt{{x}^{2} - 1}} = \sqrt{{x}^{2} - 1}$

And combining our results, we get:

$I = \frac{1}{2} {x}^{2} \setminus \text{arcsec} \setminus x - \frac{1}{2} \sqrt{{x}^{2} - 1} + c$