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 pm
Is there any way to calculate this integral on interval where the upper limit is x > a?
Titanic2001
May 5 @ 7:46 pm
No 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 pm
please solve this question integration of X/ √a^4-x^4
evo
March 23 @ 10:01 pm
Notice 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 pm
can 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 am
In 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 am
We 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)