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}

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