# How to find the antiderivative of the $\sqrt{4-x^2}$?

How to find the $$\int \sqrt{4-x^2}$$

I know that you can use substitution for x and sub in $$2\sin\theta$$, but i don't understand how and why you know to do that? That is not something I would think of doing

• Do you mean find $\int \sqrt{4-x^2}\ dx$? – Landuros Dec 24 '19 at 11:45
• @Landuros yes! sorry. – user639649 Dec 24 '19 at 11:47

$$x=2\sin\theta\implies\sqrt{4-(2\sin\theta)^2}=2\sqrt{1-\sin^2\theta}=2\cos\theta$$
$$dx=2\cos\theta\:d\theta$$ $$\int\sqrt{4-x^2}\:dx=\int2\cos\theta(2\cos\theta\:d\theta)$$ I think rest of work is easy enough. Why choosing $$x=2\sin\theta$$ is good because it convert your $$\sqrt{4-x^2}$$ into $$2\cos\theta$$. You have to remember that because of $$4$$ you need something before $$\sin\theta$$ which complete the identity $$1-\sin^2\theta$$. Hence in general $$\sqrt{a-x^2}\implies x=\sqrt{a}\sin\theta$$

The "how you do it" is in the answer from @emonHR . There are several ways to think about "why".

• The function suggests the way you rewrite the identity $$\sin^2 x + \cos^2 x = 1$$ to find $$\cos x$$ in terms of $$\sin x$$.
• You recognize the function as solving for $$y$$ in terms of $$x$$ in the equation of a circle, so it's no surprise that trigonometry may help.

• You've practiced integration for a while so you're familiar with this trick. Soon it becomes a technique, not a trick, and you have a good intution for when it will work.

Here is a geometric approach. If you let $$y=\sqrt{4-x^2}$$ then $$x^2+y^2=4$$ so the point $$(x,y)$$ lies on a circle centred on the origin with radius $$2$$.

The definite integral $$\int^a_0 y \space dx$$ is the area of the region bounded by the circle, the x axis, and the lines $$x=0$$ and $$x=a$$.

If you divide this region into a segment of the circle with angle $$\theta$$ and a right triangle with sides $$a=2\sin \theta$$ and $$\sqrt {4-a^2}=2\cos \theta$$ then you have

$$\int_0^a y \space dx = 2 \theta + 2 \sin \theta \cos \theta= 2\theta + \sin 2\theta$$.