# How to calculate $\int \sqrt{4-x^2} dx$ [closed]

I'm really confused of how to do this integral, sorry if this is a basic question:

$$\int \sqrt{4-x^2} \, dx$$

Any help would be appreciated!

• Have you learned about trigonometric substitution? Commented Nov 3, 2014 at 0:08
• A long time ago, but I have completely forgotten it Commented Nov 3, 2014 at 0:08
• Do you want the indefinite integral, or the definite integral from $-2$ to $2$? Because one would require trigonometric substitution, while the other is easily seen to equal $2\pi$. Commented Nov 3, 2014 at 0:09
• en.wikipedia.org/wiki/Trigonometric_substitution
– user9464
Commented Nov 3, 2014 at 0:13
• This question appears to be off-topic because it can be solved following the instructions in Wikipedia or by plugging it into Wolfram Alpha. Commented Nov 3, 2014 at 8:02

Use $x=2\sin u$. Do you see how it simplifies to $4\cos ^2u \,du$?

This is a classic "Trig-sub" problem. Draw yourself a right triangle with hypotenuse length $2$ and a vertical leg of length $x$. Using the Pythagorean theorem, it is clear that the length of the remaining horizontal leg must be $\sqrt{4-x^2}$. Let $\theta$ be the angle made by the hypotenuse and the leg of length $\sqrt{4-x^2}$. Then you should be able to use some basic trig to deduce that $$2\cos(\theta) = \sqrt{4-x^2}$$ and $$x = 2\sin(\theta)\implies dx = 2\cos(\theta)d\theta$$ Hence if you plug this into your integral you can rewrite it as $$\int \sqrt{4-x^2}dx =\int \left(\sqrt{4-x^2}\right)\left(dx\right) \\= \int \left(2\cos(\theta)\right)\left(2\cos(\theta)d\theta \right) \\ = \int 4\cos^2(\theta) d\theta$$ This equivalent integral is much more readily solvable with the identity $$\cos^2(x) = \frac{\cos(2x)+1}{2}$$ After you integrate, make sure to convert your solution from an equation in terms of $\theta$ back to one in terms of $x$ by making the appropriate trigonometric substitutions.