Definite Integral Inverse Tangent from 0 to Pi over 4

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

The integration of the form

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 first function {\tan ^{ - 1}}x and second function will be 1, using 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, we have

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}

We 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}