# Why can't I do this substitution?

I'm trying to solve this integral: $\int \sqrt{4 - 16x^2}$. As always I tried substitution but for some odd reason it's not working. Here's what I did:

$$u = 4 - 16x^2$$

$$\int u^\frac{1}{2} du= \frac{u^\frac{3}{2}}{\frac{3}{2}}$$

=

$$2*\frac{u^\frac{3}{2}}{3} = 2*\frac{(4-16x^2)^\frac{3}{2}}{3}$$

What's wrong with my solution?

• So ${\rm du}/{\rm dx}$ is what? Dec 18, 2017 at 19:44
• $du = -16 \cdot 2x dx$ so you can't replace $dx$ with just $du$ Dec 18, 2017 at 19:46

If you let $u=4-16x^2$ then $\frac{du}{dx}=-32x$ which you don't have multiplying outside of $\sqrt{4-16x^2}$. So best thing to do in this case is to use trig-sub. Since $\sqrt{4-16x^2}= 2\sqrt{1-4x^2}$,

let $x=\frac{1}{2}\sin t$. Then ${dx}=\frac{1}{2}\cos t dt$.

So the whole thing inside the integral turns into $2(\sqrt{1-cos ^2t}) (\frac{1}{2}\cos t) dt=\sin t \cos t dt$. You get

$\int\sqrt{4-16x^2}dx= \int 2\sqrt{1-4x^2}dx=\int \sin t\cos t dt$

Now a simple u-sub $(u=cost)$ is all that is needed.

better is to set $$x=\frac{1}{2}\sin(t)$$

Your solution misses the $du=-32xdx$, but this single $x$ cannot be substituted by some nice function of $u$.

• what can I use then?
– Trey
Dec 18, 2017 at 19:54