integration of sec inverse x

Integration of sec inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it

\[ \int \sec^{-1}x \, dx = x \sec^{-1}x – \ln |x + \sqrt { x^2- 1}| + C \]

where (C) is the constant of integration.

Proof of integration of sec inverse x

We can prove it using integration by parts and integration by substitution

Integration by parts

To find the integral we use integration by parts. Integration by parts is based on the product rule for differentiation and is given by:

$\int f(x) g(x) dx = f(x) (\int g(x) dx )- \int \left \{ \frac {df(x)}{dx} \int g(x) dx \right \} dx $
In our case, we can let $ f(x) = \sec ^{-1}x $ and $ g(x) = 1 $. Then

• $ \frac {df(x)}{dx} = \frac{1}{x\sqrt {x^2-1}} \, dx $
• $\int g(x) dx = \int dx = x $

Now, substitute these into the integration by parts formula:

$$
\int \sec^{-1}x \, dx = x \sec^{-1}x – \int x \cdot \frac{1}{x\sqrt {x^2-1}} \, dx
$$

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

Now lets calculate the second integral separately $\int \frac{1}{\sqrt {x^2 -1}} \, dx $

We will use the formula

$\int \frac {1}{\sqrt {x^2 – a^2}} dx = ln |x + \sqrt {x^2 – a^2}| + C$
Proof
Put $x =a sec \theta$ then $dx= a sec \theta tan \theta d\theta $
$\int \frac {1}{\sqrt {x^2 -a^2}} dx $
$=\int sec \theta d\theta $
$=ln |sec \theta + tan \theta| + C $
Now
$sec \theta = \frac {x}{a} $
$tan \theta = \sqrt {sec^2 \theta -1 } = \sqrt {\frac {x^2}{a^2} -1} $
Therefore
$=ln | \frac {x}{a} + \sqrt {\frac {x^2}{a^2} -1}| + C_1 = ln |x + \sqrt {x^2 – a^2}| – ln |a| +C_1 = ln |x + \sqrt {a^2 + x^2}| + C$

Substituting this value in main integral , we get

$$
\int \sec^{-1}x \, dx = x \sec^{-1}x – \ln |x + \sqrt { x^2+ 1}| + C
$$

Integration by substitution

let $\theta = \sec^{-1}x$
$ \sec \theta = x$
$\sec \theta tan \theta d \theta = dx$

$$
\int \sec^{-1}x \, dx = \int \theta (\sec \theta tan \theta) d \theta
$$

Now solving this using integration parts

$ = \theta \int (\sec \theta tan \theta) d \theta – \int (\frac {d}{d \theta } \theta ) .(\int (\sec \theta tan \theta)) \; \theta \\
= \theta \sec \theta – \int \sec \theta \; d \theta $

Now

\[
\int \sec(x) \, dx = \ln | \sec(x) + \tan(x) | + C
\]

Therefore

$=\theta \sec \theta – \ln | \sec(x) + \tan(x) | + C$

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

Substituting back the values

$\int \sec^{-1}x \, dx = x \sec^{-1}x – \ln |x + \sqrt {x^2 -1}|+ C$

Solved Examples on integration of sec inverse x

Question 1

integral of $x \sec^{-1}x $

Solution

Applying integration by parts
$\int u \, dv = uv – \int v \, du$.
Let $u = \sec^{-1}x$. Then, $du = \frac{1dx}{x\sqrt{ x^2-1}}$.
Let $dv = x \, dx$. Then, $v = \frac{x^2}{2}$.
Then
$ \int x \sec^{-1}x \, dx = \frac{x^2}{2} \sec^{-1}x – \int \frac{x^2}{2} \cdot \frac{1dx}{x\sqrt{x^2 -1}} \
=\frac{x^2}{2} \sec^{-1}x – \frac {1}{2} \int \frac{x}{\sqrt {x^2-1}} \; dx$
Now lets solve the integral separately
$\int \frac{x}{\sqrt{x^2 -1}} \; dx$

Let $x^2 -1 = t$
$2x dx= dt$
therefore
$\int \frac{x}{\sqrt{ x^2 -1}} \; dx =\frac {1}{2} \int t^{-1/2} \; dt$
$= t^{1/2} = \sqrt{x^2 -1}$

Applying these steps to the integral of $x \sec^{-1}x$, we obtained:

$$\frac{x^2}{2} \sec^{-1}x – \frac {\sqrt{x^2 -1}}{2}+ C $$

where $C$ is the constant of integration.