Integration of Square Root of a^2+x^2

In this tutorial we shall drive the integration of square root of a^2+x^2, and solve this integration with the help of integration by parts methods.

The integral of \sqrt {{a^2} + {x^2}} is of the form

\int  {\sqrt {{a^2} + {x^2}} dx}  =  \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sinh ^{ - 1}}\left(  {\frac{x}{a}} \right) + c


OR

\int  {\sqrt {{a^2} + {x^2}} dx}  =  \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}\ln \left[ {x + \sqrt  {{a^2} + {x^2}} } \right] + c


This integral can be written as

I =  \int {\sqrt {{a^2} + {x^2}}  \cdot 1dx}


Here first function is \sqrt  {{a^2} + {x^2}} and second function will be 1

I =  \int {\sqrt {{a^2} + {x^2}}  \cdot 1dx}  \,\,\,\,{\text{ -  -  - }}\left( {\text{i}} \right)


Using formula for integration by parts, we have

\int  {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int  {g\left( x \right)dx - \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int  {g\left( x \right)dx} } \right]dx} } }


Equation (i) becomes using above formula, we have

\begin{gathered} I = \sqrt {{a^2} + {x^2}} \int {1dx - \int  {\left[ {\frac{d}{{dx}}\sqrt {{a^2} + {x^2}} \left( {\int {1dx} } \right)}  \right]} dx} \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}}  - \int {\frac{{ - {a^2} + {a^2} +  {x^2}}}{{\sqrt {{a^2} + {x^2}} }}} dx \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}}  - \int {\frac{{ - {a^2}}}{{\sqrt {{a^2} +  {x^2}} }}} dx - \int {\frac{{{a^2} + {x^2}}}{{\sqrt {{a^2} + {x^2}} }}dx} \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}}  + {a^2}\int {\frac{1}{{\sqrt {{a^2} + {x^2}}  }}} dx - \int {\sqrt {{a^2} + {x^2}} dx} \\ \end{gathered}


Putting the given integral I = \int {\sqrt {{a^2} + {x^2}} } dx, we get

\begin{gathered} I = x\sqrt {{a^2} + {x^2}}  + {a^2}{\sinh ^{ - 1}}\left( {\frac{x}{a}}  \right) - I + c \\ \Rightarrow I + I = x\sqrt {{a^2} +  {x^2}}  + {a^2}{\sinh ^{ - 1}}\left(  {\frac{x}{a}} \right) + c \\ \Rightarrow 2I = x\sqrt {{a^2} + {x^2}}  + {a^2}{\sinh ^{ - 1}}\left( {\frac{x}{a}}  \right) + c \\ \Rightarrow I = \frac{{x\sqrt {{a^2} +  {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sinh ^{ - 1}}\left( {\frac{x}{a}} \right) + c \\ \Rightarrow \int {\sqrt {{a^2} + {x^2}}  dx}  = \frac{{x\sqrt {{a^2} + {x^2}}  }}{2} + \frac{{{a^2}}}{2}{\sinh ^{ - 1}}\left( {\frac{x}{a}} \right) + c \\ \end{gathered}


But using the relation {\sinh  ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right) and we know that {\sinh ^{ - 1}}\left( {\frac{x}{a}}  \right) = \ln \left( {x + \sqrt {{x^2} + {a^2}} } \right) we have

  \Rightarrow \int {\sqrt {{a^2} + {x^2}} dx}   = \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}\ln \left( {x +  \sqrt {{x^2} + {a^2}} } \right) + c

Comments

comments