How does Wolfram Alpha come up with the substitution $x = 2\sin u$? Integration/Analysis

Question

I have to integrate $$ \int_0^2 \sqrt{4-x^2} \, dx $$

I looked at the Wolfram Alpha step by step solution, and first thing it does is it substitutes

$x = 2\sin(u)\text{ and } \,dx = 2\cos(u)\,du$

How does it know to substitute $2\sin(u)$ for $x$?

Answer 1

This is based on the fact that the usual parametric equation of the curve $x^2 + y^2 = R^2$ is $X = R\sin t, Y = R\cos t$.

You want to make a change of variables, to make the integrand look simpler (that is the point of the change of variables). So you would like to simplify $$ \sqrt{4 - x^2} $$and in particular get rid of the $\sqrt.$. So you look for a function $t$ such as $$ y(t)^2 = 4-x(t)^2 $$

has a simple form. And you get $$ x(t) = 2\sin t, y(t) = 2\cos t $$ for the right domain of integration.

Answer 2

This is a common substitution technique known as Trig substitution, you substitute $x$ with a trigonometric function.

Typically, when something is in the form $\sqrt{a-x^2}$, you substitute $x=\sqrt a\sin u$

Answer 3

Theóphile's comment is a good answer. I will give another.

The $\sqrt{4-x^2}$ and indeed anything of the form $\sqrt{a^2-b^2}$ --- think $\sqrt{x^2-b^2}$ or $\sqrt{a^2-x^2}$ --- or $x^2+b^2$ can be considered --- via Pythagoras --- as sides of a right-angled triangle.

In this example we have $\sqrt{2^2-x^2}=b\Rightarrow b^2+x^2=2^2$ so that we have a right-angled-triangle with sides $x$, $b=\sqrt{2^2-x^2}$ and hypotenuse $2$.

Draw this triangle. Now choose an angle $\theta$ in the triangle so that you can write

$$\sec/\tan/\sin\theta=\frac{x}{2}.$$

In this case we have $\sin \theta=\frac{x}{2}$.

Now, using the triangle you sketched, $$\cos\theta=\frac{\sqrt{4-x^2}}{2}\Rightarrow \sqrt{4-x^2}=2\cos\theta.$$

We can also get a handle on $dx$ no problem.

When we eventually antidifferentiate, our answer will be in terms of $\theta$. We can find any of the trig ratios in terms of $x$ using our triangle and $\theta$... well $\theta$ is than angle whose sine is $\frac{x}{2}$ so $$\theta=\sin^{-1}\left(\frac{x}{2}\right).$$ ... not that it arises in this question.