The integral of $\sqrt{1-x^2}$ is equal to $\dfrac{x}{2} \sqrt{1-x^2}$ $+\dfrac{1}{2}$ sin^{-1} x+C. Here we will learn how to integrate square root of 1-x^{2}.

The integral formula of root 1-x^2 is provided below: $\int \sqrt{1-x^2} dx$ $=\dfrac{x}{2} \sqrt{1-x^2}$ $+\dfrac{1}{2}$ sin ^{-1} x+C. |

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

**Question:** How to Integrate Root (1-x^2)?

That is, find $\int \sqrt{1-x^2} dx$.

**Solution:**

Let x = sint.

So dx = cost dt

Also, we have

- sin2t = 2sint cost = 2sint $\sqrt{1-\sin^2t}$ = 2x$\sqrt{1-x^2}$ …(∗)
- t = sin
^{-1}x …(∗∗)

Now, $\int \sqrt{1-x^2} dx$

= $\int \sqrt{1-\sin^2t} \cos t \,dt$

= $\int \cos^2 t \, dt$ as we know cos^{2}θ=1-sin^{2}θ

= $\dfrac{1}{2}\int 2\cos^2 t \,dt$

= $\dfrac{1}{2}\int (1+\cos 2t) dt$ as 1+cos2θ = 2cos^{2}θ.

= $\dfrac{1}{2} \big(t+\dfrac{\sin 2t}{2} \big)$+C

= $\dfrac{1}{2} \big(\sin^{-1}x+\dfrac{2x\sqrt{1-x^2}}{2} \big)$+C, using Equations (∗) and (∗∗).

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

So the integration of $\sqrt{1-x^2}$ is equal to $\int \sqrt{1-x^2} dx$ $=\dfrac{x}{2} \sqrt{1-x^2}$ $+\dfrac{1}{2}$ sin^{-1} x+C, where C is an integration constant.

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

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

## FAQs

### Q1: What is the integration of sqrt{1-x^2}?

Answer: The integration of square root of 1-x^2 is equal to $\dfrac{x}{2} \sqrt{1-x^2}$ $+\dfrac{1}{2}$ sin^{-1} x+C, where C is an arbitrary constant.