Integration of Square Root (a^2-x^2) | Integral of Root (a^2-x^2)

In this post, we will find the integral of root(a²-x²). The integration of square root of a²-x² is given as follows

∫ √(a²-x²) dx = x/2 ⋅ √(a²-x²) + a²/2 sin^-1(x/a) + C, where C is an integration constant.

Integration of $\sqrt{a^2-x^2}$

Let $I = \int \sqrt{a^2-x^2} dx$ $\cdots (I)$

We will find the integration of the square root of a²-x² using the integration by parts formula. The formula says that if f(x), g(x) are two functions then the integration of f(x)g(x) is given by

$\int f(x)g(x) dx$ $=f \int g dx$ $-\int \big[\dfrac{df}{dx} \int g dx \big]dx$ $\cdots (\star)$

Put $f(x)=\sqrt{a^2-x^2}$ and $g(x)=1$.

Now,

∫g(x) dx = ∫1 dx = x

and

$\dfrac{d}{dx}(f(x))=\dfrac{d}{dx}(\sqrt{a^2-x^2})$ $=\dfrac{d}{dx}((a^2-x^2)^{1/2})$ $=\dfrac{1}{2}(a^2-x^2)^{1/2-1}\cdot \dfrac{d}{dx}(a^2-x^2)$ by the chain rule and power rule of derivatives. $=\dfrac{-x}{\sqrt{a^2-x^2}}$

Now, using the formula $(\star)$, the integral of root(a^2-x^2) is

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

= $\sqrt{a^2-x^2} \cdot x$ $-\int \dfrac{-x}{\sqrt{a^2-x^2}} \cdot x dx$

= $x\sqrt{a^2-x^2}$ $-\int \dfrac{a^2-x^2-a^2}{\sqrt{a^2-x^2}}$

= $x\sqrt{a^2-x^2}-\int \sqrt{a^2-x^2} dx$ $+a^2 \int \dfrac{dx}{a^2-x^2}$

Therefore,

I = $x\sqrt{a^2-x^2}-I$ $+a^2 \int \dfrac{dx}{a^2-x^2}$

⇒ 2I = $x\sqrt{a^2-x^2}$ $+a^2 \int \dfrac{dx}{a^2-x^2}$

⇒ 2I = $x\sqrt{a^2-x^2}$ $+a^2 \sin^{-1} \dfrac{x}{a}+C$

∴ The integral of root a²-x² is equal to $\dfrac{x}{2} \sqrt{a^2-x^2}$ $+\dfrac{a^2}{2}\sin^{-1} \dfrac{x}{a}+C$, where C is an integration constant.

Application

Using the above formula, we can easily find the integral of square root of 4-x².

Put a=2 in the above formula.

So the integration of root 4-x² is equal to

∫ $\sqrt{4-x^2} dx$ $=\dfrac{x}{2} \sqrt{4-x^2}$ +2 sin^-1(x/2) +C.

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