# Undoing the trig substitution

I was given the following problem to integrate: $$\int\frac{x}{\sqrt{4x-x^2}}\ dx$$ My instinct is to begin by drawing a triangle:

We know can create the following identities: $$\tan(\theta)=\frac{x}{\sqrt{4x-x^2}}\\\sin(\theta)=\frac{x}{2\sqrt{x}}\\\cos(\theta)=\frac{\sqrt{4x-x^2}}{2\sqrt{x}}$$

We can now rewrite the integral: $$\int\tan(\theta)\ dx$$ Our remaining problem is the dx. We can differentiate our sine identity to help us with this: $$\frac{d}{dx}\sin(\theta)=\frac{d}{dx}\frac{x}{2\sqrt{x}}\\=\cos\theta\frac{d\theta}{dx}=\frac{1}{4\sqrt{x}}\\=4\sqrt{x}\cos\theta \ d\theta=dx$$ Now we need to find the ralationship between theta and x. Using the sine identity again: $$\sin(\theta)=\frac{x}{2\sqrt{x}}\\\sin^2(\theta)=\frac{x}4\\=4\sin^2(\theta)=x$$ Our rewritten integral can now be: $$\int\tan(\theta)\ 4\sqrt{(4\sin^2(\theta)}\cos(\theta) \ d\theta\\=\int\sin(\theta)\ 8\sin(\theta) \ d\theta\\=\int8\sin^2(\theta)\ d\theta\\=8\int\frac{1-\cos(2\theta)}{2}\ d\theta\\=4\left(\theta-\frac{\sin(2\theta)}{2}\right)$$My problem is now undoing my original substitution. I don't know how to deal with the $$\sin(2\theta)$$ because of the two. I am also unsure of how to take care of the $$\theta$$ term. Solving for $$\theta$$ will give me an arcsine or arccos or arctan... How do I undo this substitution?

• Well, if $\sin(\theta) = \frac{x}{2\sqrt{x}} = \frac{1}{2}\sqrt{x}$, then $\theta=\arcsin(\frac{1}{2}\sqrt{x})$. As for $\sin(2\theta)$, you can use the addition formula $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and the values you already have for them. – Arturo Magidin Dec 5 '19 at 20:45

$$\sin\theta=\frac{\sqrt{x}}{2},\>\>\>\>\cos\theta=\frac{\sqrt{4x-x^2}}{2\sqrt{x}}$$
$$\theta = \sin^{-1}\frac{\sqrt{x}}{2}$$ $$\sin 2\theta = 2\sin\theta\cos\theta = \frac12\sqrt{4x-x^2}$$
$$I=4\left(\theta-\frac{\sin2\theta}{2}\right)=4\sin^{-1}\frac{\sqrt{x}}{2}-\sqrt{4x-x^2}$$