integration of sec cube x

The integration of sec cube x $\sec ^3 x$, can be found using integration substitution and trigonometry identities . The integral of $\sec ^3 x$ with respect to (x) is:

\[
\int \sec^3 x \, dx =\frac {1}{2} \sec x \tan x +\frac {1}{2} \ln |sec x + tan x|+ C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

Proof of integration of sec cube x

To find the integral of sec cube x, $\sec ^3 x$, 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(x) $ and $ g(x) = \sec^2x $. Then

• $ \frac {df(x)}{dx} = \sec x tan x \, dx $
• $\int g(x) dx =\tan x $

Substituting these values, we get

$I=\int \sec^3 x \, dx = \sec x \tan x – \int \sec x \tan x \tan x dx$
$I= \sec x \tan x – \int \tan^2 x \sec x dx$
$I= \sec x \tan x – \int (\sec^2 x – 1) \sec x dx$
$I= \sec x \tan x – \int \sec^3 x \, dx + \int \sec x dx
$2I=\sec x \tan x + \ln |sec x + tan x| + C$

or

$\int \sec^3 x \, dx =\frac {1}{2} \sec x \tan x +\frac {1}{2} \ln |sec x + tan x|+ C$

The integral of (\sec^3(x)) is known for being more complex than simpler trigonometric functions

Solved examples

Question

Calculate the definite integral of $ \sec^3(x) $ from ( x = 0 ) to $ x = \frac{\pi}{4} $.

Solution:

We need to evaluate:

\[
\int_0^{\frac{\pi}{4}} \sec^3(x) \, dx
\]

From the earlier calculation, we know that the indefinite integral of ( \sec^3(x) ) is:

$\int \sec^3 x \, dx =\frac {1}{2} \sec x \tan x +\frac {1}{2} \ln |sec x + tan x|+ C$

Now $ \tan 0 =0 $ , $\sec 0 =1 $
$\tan \pi/4 = 1$, $ \sec \pi/4 = \sqrt 2$

Substituting these , we get

\[
\int_0^{\frac{\pi}{4}} \sec^3(x) \, dx = \frac {1}{\sqrt 2} + \frac {1}{2} \ln |\sqrt 2 + 1|
\]