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

Mar 3, 2018

The answer is $= x \text{arc} \sec x - \ln \left(x + \sqrt{{x}^{2} - 1}\right) + C$

Explanation:

We need

$\left({\sec}^{-} 1 x\right) ' = \left(\text{arc} \sec x\right) ' = \frac{1}{x \sqrt{{x}^{2} - 1}}$

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

$\int u ' v = u v - \int u v '$

Here, we have

$u ' = 1$, $\implies$, $u = x$

$v = \text{arc} \sec x$, $\implies$, $v ' = \frac{1}{x \sqrt{{x}^{2} - 1}}$

Therefore,

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

Perform the second integral by substitution

Let $x = \sec u$, $\implies$, $\mathrm{dx} = \sec u \tan u \mathrm{du}$

$\sqrt{{x}^{2} - 1} = \sqrt{{\sec}^{2} u - 1} = \tan u$

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

$= \int \frac{\sec u \left(\sec u + \tan u\right) \mathrm{du}}{\sec u + \tan u}$

$= \int \frac{\left({\sec}^{2} u + \sec u \tan u\right) \mathrm{du}}{\sec u + \tan u}$

Let $v = \sec u + \tan u$, $\implies$, $\mathrm{dv} = \left({\sec}^{2} u + \sec u \tan u\right) \mathrm{du}$

So,

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 1}} = \int \frac{\mathrm{dv}}{v} = \ln v$

$= \ln \left(\sec u + \tan u\right)$

$= \ln \left(x + \sqrt{{x}^{2} - 1}\right)$

Finally,

$\int \text{arc"secxdx=x"arc} \sec x - \ln \left(x + \sqrt{{x}^{2} - 1}\right) + C$

Mar 3, 2018

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

Explanation:

Alternatively, we can use a little-known formula for working out integrals of inverse functions. The formula states:
$\int \setminus {f}^{-} 1 \left(x\right) \setminus \mathrm{dx} = x {f}^{-} 1 \left(x\right) - F \left({f}^{-} 1 \left(x\right)\right) + C$
where ${f}^{-} 1 \left(x\right)$ is the inverse of $f \left(x\right)$ and $F \left(x\right)$ is the anti-derivative of $f \left(x\right)$.

In our case, we get:
$\int \setminus {\sec}^{-} 1 \left(x\right) \setminus \mathrm{dx} = x {\sec}^{-} 1 \left(x\right) - F \left({\sec}^{-} 1 \left(x\right)\right) + C$

Now all we need to work out is the anti-derivative $F$, which is the familiar secant integral:
$\int \setminus \sec \left(x\right) \setminus \mathrm{dx} = \ln | \sec \left(x\right) + \tan \left(x\right) | + C$

Plugging this back into the formula gives our final answer:
$\int \setminus {\sec}^{-} 1 \left(x\right) \setminus \mathrm{dx} = x {\sec}^{-} 1 \left(x\right) - \ln | \sec \left({\sec}^{-} 1 \left(x\right)\right) + \tan \left({\sec}^{-} 1 \left(x\right)\right) | + C$

We need to be careful about simplifying $\tan \left({\sec}^{-} 1 \left(x\right)\right)$ to $\sqrt{{x}^{2} - 1}$ because the identity is only valid if $x$ is positive. We are lucky, however, because we can fix this by putting an absolute value on the other term inside the logarithm. This also removes the need for the first absolute value, since everything inside the logarithm will always be positive:
$x {\sec}^{-} 1 \left(x\right) - \ln \left(| x | + \sqrt{{x}^{2} - 1}\right) + C$