# 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

\[I = \int {\sqrt {{a^2} – {x^2}} dx} = \frac{{x\sqrt {{a^2} – {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ – 1}}\left( {\frac{x}{a}} \right) + c\]

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

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

MK

September 18@ 6:40 pmIs there any way to calculate this integral on interval where the upper limit is x > a?

Titanic2001

May 5@ 7:46 pmNo because the function itself will be undefined after x=a. And you can’t find the integral of a undefined function can you?

Bipin

March 4@ 9:58 pmplease solve this question integration of X/ √a^4-x^4

evo

March 23@ 10:01 pmNotice that x^4 is (x^2)^2 whose derivative is 2x (which is in the numerator if you multiply and divide the integral by 2). Now it’s basically another integral like the one in the article

neil gayo

October 29@ 10:22 pmcan i get the answer of this ? y=√a2+x2

karina bohora

February 17@ 3:41 pm√2-x-√2+x/x

Marc

December 7@ 4:00 amIn line 4 you have (a^2 – x^2) / (sqrt(x^2 – a^2)) in the integral as the last addend

In the next line something disappears and you have only sqrt(x^2 – a^2) in the integral

Can you explain how you got from line 4 to line 5, I think you made a mistake there

eMathZone

December 9@ 2:03 amWe can write a = (sqrt(a))^2, so (a^2 – x^2) = (sqrt(x^2 – a^2))^2 so this is what we get: (a^2 – x^2) / (sqrt(x^2 – a^2)) = (sqrt(a^2 – x^2))^2 / (sqrt(x^2 – a^2)) = sqrt(x^2 – a^2)