Integration of Root (a^2-x^2) | Integration of Root (1-x^2)

In this post, we will find the integral of root($a^2-x^2$). The integration of the square root of $a^2-x^2$ is given below:

$\int \sqrt{a^2-x^2} dx$ $=\dfrac{x}{2} \sqrt{a^2-x^2}$ $+\dfrac{a^2}{2}\sin^{-1} \dfrac{x}{a}+C$, 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 $a^2-x^2$ 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

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

In this formula $(\star)$, we put $f(x)=\sqrt{a^2-x^2}$ and $g(x)=1$.

Now, $\int g(x) dx$ $=\int 1 dx =x$

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 integration.

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

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,

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

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

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

$\therefore \int \sqrt{a^2-x^2} dx$ $=\dfrac{x}{2} \sqrt{a^2-x^2}$ $+\dfrac{a^2}{2}\sin^{-1} \dfrac{x}{a}+C$, where $C$ is an integration constant.

Also Read: 

Integration of log(sinx) from 0 to pi/2

Integration of e^3x

Derivative & integration of 1/root(x)

Integration of 1/(1+x²)

Integration of $\sqrt{1-x^2}$

Question: Find the integral $\int \sqrt{1-x^2} dx$.

Answer:

Put $a=1$ in the above formula. Then the integration of $\sqrt{1-x^2}$ will be equal to

$\int \sqrt{1-x^2} dx$ $=\dfrac{x}{2} \sqrt{1-x^2}$ $+\dfrac{1}{2}\sin^{-1} \dfrac{x}{1}+C$

$=\dfrac{x}{2} \sqrt{1-x^2}$ $+\dfrac{1}{2}\sin^{-1} x+C$.

Here C is an integral constant.

FAQs