The Definite Integral of Tanx from 0 to Pi over 4

In this tutorial we shall derive the definite integral of the trigonometric function tangent from limits 0 to Pi over 4.

The integration of the form is
$I = \int\limits_0^{\frac{\pi }{4}} {\tan xdx}$

First we evaluate this integration by using the integral formula $\int {\tan xdx = – \ln \cos x}$, and then we use the basic rule of the definite integral $\int\limits_a^b {f\left( x \right)dx = \left| {F\left( x \right)} \right|_a^b} = \left[ {F\left( b \right) – F\left( a \right)} \right]$. So we have
$\begin{gathered} \int\limits_0^{\frac{\pi }{4}} {\tan xdx} = \left| { – \ln \cos x} \right|_0^{\frac{\pi }{4}} \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {\tan xdx} = – \left| {\ln \cos x} \right|_0^{\frac{\pi }{4}} \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {\tan xdx} = – \left[ {\ln \cos \frac{\pi }{4} – \ln \cos 0} \right] \\ \end{gathered}$

From the trigonometric values $\cos 0 = 1$ and $\cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$, we have
$\Rightarrow \int\limits_0^{\frac{\pi }{4}} {\tan xdx} = – \left[ {\ln \frac{1}{{\sqrt 2 }} – \ln 1} \right]$

Using the value $\ln 1 = 0$, we have
$\begin{gathered} \Rightarrow \int\limits_0^{\frac{\pi }{4}} {\tan xdx} = – \left[ {\ln \frac{1}{{\sqrt 2 }} – 0} \right] \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {\tan xdx} = \ln {\left( {\frac{1}{{\sqrt 2 }}} \right)^{ – 1}} = \ln \sqrt 2 \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {\tan xdx} = \ln {2^{\frac{1}{2}}} = \frac{1}{2}\ln 2 \\ \end{gathered}$