Integration of the Square Root of a^2+x^2
In this tutorial we shall derive the integration of the square root of a^2+x^2, and solve this integration with the help of the 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 the first function is $$\sqrt {{a^2} + {x^2}} $$ and the second function is $$1$$
\[I = \int {\sqrt {{a^2} + {x^2}} \cdot 1dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]
Using the formula for integration by parts, we have
\[\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx – \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } \]
Using the formula above, equation (i) becomes
\[\begin{gathered} I = \sqrt {{a^2} + {x^2}} \int {1dx – \int {\left[ {\frac{d}{{dx}}\sqrt {{a^2} + {x^2}} \left( {\int {1dx} } \right)} \right]} dx} \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}} – \int {\frac{{ – {a^2} + {a^2} + {x^2}}}{{\sqrt {{a^2} + {x^2}} }}} dx \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}} – \int {\frac{{ – {a^2}}}{{\sqrt {{a^2} + {x^2}} }}} dx – \int {\frac{{{a^2} + {x^2}}}{{\sqrt {{a^2} + {x^2}} }}dx} \\ \Rightarrow I = x\sqrt {{a^2} + {x^2}} + {a^2}\int {\frac{1}{{\sqrt {{a^2} + {x^2}} }}} dx – \int {\sqrt {{a^2} + {x^2}} dx} \\ \end{gathered} \]
Using 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 since we know that $${\sinh ^{ – 1}}\left( {\frac{x}{a}} \right) = \ln \left( {x + \sqrt {{x^2} + {a^2}} } \right)$$, we have
\[ \Rightarrow \int {\sqrt {{a^2} + {x^2}} dx} = \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + \frac{{{a^2}}}{2}\ln \left( {x + \sqrt {{x^2} + {a^2}} } \right) + c\]
umit
July 15 @ 9:27 pm
why sinh^-1 (x/a)= In (x+sqrt(x^2+a^2) ? I found In(x+sqrt(x^2+a^2)/a)
yCombinator
June 19 @ 9:45 am
It is not the value of sin^-1(x/a) nor (1/a).tan^-1(x/a) that you are confused about, rather an algebraic problem in itself, to solve which, you make use of trignometric substitutions which you might have learned earlier. basically you put x = atan(x) for the form int(sqrt(a^2 ^x^2)) and simply the expression to a form we can use basic proven trig functions’ integrals. Intuition of pythagorean theorem is being comes handy here!