0
$\begingroup$

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:
enter image description here

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Dec 5 '19 at 20:45
0
$\begingroup$

Use the identities identified earlier,

$$\sin\theta=\frac{\sqrt{x}}{2},\>\>\>\>\cos\theta=\frac{\sqrt{4x-x^2}}{2\sqrt{x}}$$

to undo the angles,

$$\theta = \sin^{-1}\frac{\sqrt{x}}{2}$$ $$\sin 2\theta = 2\sin\theta\cos\theta = \frac12\sqrt{4x-x^2}$$

Thus, the integral becomes,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.