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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

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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