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Integration of tan x

Integration of tan x can be found using various integration technique like integration by substitution along with trigonometric identities. The formula for integration of tan x is

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

Proof of the Integration of tan x

Integration of tan x can be solved using integration by substitution as given below

$\int \frac {f^{‘} (x)}{f(x)} \; dx = ln | f(x)| + C$
Proof
let us substitute
f(x) =t
then
f'(x) dx=dt
Therefore
$\int \frac {f^{‘} (x)}{f(x)}\; dx = \int \frac {1}{t} \; dt =ln |t| + C= ln | f(x)| + C$

Now

$\int \tan(x) \, dx = \int \frac {\sin(x)}{\cos(x)} \, dx$

Now
$\cos(x) =t$
then
$-\sin (x) dx = dt$
Therefore,

$\int \cot(x) \, dx =- \int \frac {1}{t} \, dt = -\ln |cos(x)| + C$

$= \ln |cos(x)|^{-1} + C=\ln |\frac {1}{\cos(x)}| + C= \ln |sec(x)| + C$

Definite Integral of tan x

To evaluate the definite integral of $\tan(x)$ over a specific interval, say from (a) to (b), you follow the same process as for the indefinite integral, but then apply the limits of integration. The indefinite integral of $\tan(x)$ is $\ln|\sec(x)| + C$.

So, the definite integral

$\int_{a}^{b} \tan(x) \, dx = \left[ \ln|\sec(x)| \right]_{a}^{b} = \ln|\sec(b)| – \ln|\sec(a)|.$

However, it’s important to be cautious about the interval of integration because $\tan(x)$ and $\ln|\sec(x)|$ have singularities (points where they are not defined). Specifically, $\tan(x)$ and $\ln|\sec(x)|$ are undefined where $\cos(x) = 0$, which occurs $\frac {\pi}{2} + k \pi$ where k is an integers. Therefore, the interval ([a, b]) should not include points where $\cos(x) = 0$

Example

$\int_{0}^{\pi/2} \tan(x) \, dx = \left[ \ln|\sec(x)| \right]_{0}^{\pi/2} = \ln|\sec(\pi/2)| – \ln|\sec(0)| = \ln( undefined) – \ln(1).$

However, this integral is problematic at ($x =\pi/2$) ) because $\ln|\sec(\pi/2)|$ is undefined (as $\sec(\pi/2)$ is not defined). In such cases, the integral does not have a standard value

Solved Examples

Question 1

Find

$\int_{0}^{\pi/4} \tan(x) \, dx$

Solution

$\int_{0}^{\pi/4} \tan(x) \, dx = \left[ \ln|\sec(x)| \right]_{0}^{\pi/4}.$

Now, evaluate this at the upper and lower limits:

\begin{align} \left[ \ln|\sec(x)| \right]_{0}^{\pi/4} = \ln|\sec(\pi/4)| – \ln|\cos(0)| \\ = \ln\left|\sqrt{2}\right| – \ln|1| \\ = \ln(\sqrt{2}). \end{align}

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