# The Definite Integral Inverse Tangent from 0 to Pi over 4

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

The integration of the form is
$I = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{ – 1}}xdx}$
$I = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{ – 1}}x \cdot 1dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Here the first function ${\tan ^{ – 1}}x$ and the second function will be $1$. Using the formula for integration by parts in definite integral form, we have
$\int\limits_a^b {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int\limits_a^b {g\left( x \right)dx – \int\limits_a^b {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)} } \right]} dx} }$

Equation (i) becomes:
$I = {\tan ^{ – 1}}x\int\limits_0^{\frac{\pi }{4}} {1dx – } \int\limits_0^{\frac{\pi }{4}} {\left[ {\frac{d}{{dx}}{{\tan }^{ – 1}}x\left( {\int {1dx} } \right)} \right]dx}$

Using the basic rule of definite integral $\int\limits_a^b {f\left( x \right)dx = \left| {F\left( x \right)} \right|_a^b} = \left[ {F\left( b \right) – F\left( a \right)} \right]$, we have
$\begin{gathered} I = \left| {x{{\tan }^{ – 1}}x} \right|_0^{\frac{\pi }{4}} – \int\limits_0^{\frac{\pi }{4}} {\left[ {\frac{x}{{1 + {x^2}}}} \right]} dx \\ \Rightarrow I = \left| {x{{\tan }^{ – 1}}x} \right|_0^{\frac{\pi }{4}} – \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\left[ {\frac{{2x}}{{1 + {x^2}}}} \right]} dx \\ \Rightarrow I = \left| {x{{\tan }^{ – 1}}x} \right|_0^{\frac{\pi }{4}} – \frac{1}{2}\left| {\ln \left( {1 + {x^2}} \right)} \right|_0^{\frac{\pi }{4}} \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{ – 1}}xdx} = \left[ {\frac{\pi }{4}{{\tan }^{ – 1}}\frac{\pi }{4} – 0} \right] – \frac{1}{2}\left[ {\ln \left( {1 + {{\left( {\frac{\pi }{4}} \right)}^2}} \right) – \ln \left( {1 + 0} \right)} \right] \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{ – 1}}xdx} = \frac{\pi }{4}{\tan ^{ – 1}}\frac{\pi }{4} – \frac{1}{2}\left[ {\ln \left( {1 + \frac{{{\pi ^2}}}{{16}}} \right) – \ln 1} \right] \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{ – 1}}xdx} = \frac{\pi }{4}{\tan ^{ – 1}}\frac{\pi }{4} – \frac{1}{2}\left[ {\ln \left( {1 + \frac{{{\pi ^2}}}{{16}}} \right)} \right] \\ \end{gathered}$