integration of sec square x

The integral of the sec square function, $\sec ^2x$, can be found using basic calculus principles. The integral of ($\sec ^2x$) with respect to (x) is:

\[
\int \sec^2 x \, dx = \tan x + C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant. This result is derived from the fact that the derivative of ($\tan x$) is $\sec^2 x$.

Integration of sec square x by Principal of Derivatives

we can use the fundamental theorem of calculus, which states that if a function f is the derivative of another function F, then the integral of f is F plus a constant.

Now we know that

$\frac {d}{dx} \tan (x)= \sec^2(x)$
or
$\frac {d}{dx} (\tan(x)+ C) = \sec^2(x)$

Where C is the constant

Now from fundamental theorem of calculus stated above, we can say that

\[
\int \sec^2 x \, dx = \tan x + C
\]

Integration of sec square x by substitution method

Let $t= \tan x$
$dt = \sec^2 x dx$
Therefore

\[
\int \sec^2 x \, dx = \int \; dt
\]

$= t + C$

Substituting back the values

\[
\int \sec^2 x \, dx = \tan x + C
\]

Solved Examples based on Integration of sec square x

Example 1

$\int x \sec^2 x \, dx$

Solution

To solve the integral , we need to use integration by parts, which is given by the formula:

$$
\int u \, dv = uv – \int v \, du
$$

Here, we can choose $ u = x $ and $ dv = \sec^2 x \, dx $. Then we need to find $ du $ and $ v $.

1. $ u = x $ implies $ du = dx $.
2. $ dv = \sec^2 x \, dx $ implies $ v = \int \sec^2 x \, dx $.

The integral of $\sec^2 x$ is $ \tan x $, so $ v = \tan x $.

Substituting these

$$
\int x \sec^2 x \, dx = x(\tan x) – \int (\tan x) \, dx
$$

Now we know that $\int \tan x \, dx= \ln |sec(x)|$

Therefore, the final value is
$$
=x \tan x + \log|\sec x| + C
$$

where $C$ is the constant of integration.

Example 2

$\int \sec^2 (2x) \, dx$

Solution

Let $t= 2 x$
$dt = 2 dx$
Therefore

$\int \sec^2 (2x) \, dx = \frac {1}{2} \int \sec^2 (t) \, dt $
$= \frac {1}{2}( \tan t + C)$

Substituting back

$\int \sec^2 (2x) \, dx = \frac {1}{2} ( \tan 2x) + C)$

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