Solve $\int\frac{1}{\sqrt{9-4x^2}}dx$

Question

I need to solve $$\int\frac{1}{\sqrt{9-4x^2}}dx$$

The first thing that came to my mind was that $9-4x^2$ can be expressed as a difference of squares, because $9-4x^2=3^2-(2x)^2=(3-2x)(3+2x)$.

This factorized form looked more promising, so I went on with it and my integral became

$$\int\frac{1}{\sqrt{(3-2x)(3+2x)}}dx$$

Here, finding no promising way to move on, I thought perhaps a substitution was in order. So I let $u=3-2x$, $du=-2dx$. From this it follows too that $3+2x=6-u$- Then $(3-2x)(3+2x)=u(6-u)$. Then I had

$$-\frac{1}{2}\int\frac{1}{\sqrt{u(u-6)}}du$$

This looks much nicer, but I have not found any way to continue from here. Going back to the beginning I find no other good ways to solve the problem. So I'm either missing another path to follow (different from factorizing and then substituting) or I did take the right approach but can't solve this last integral. Any suggestions?

Answer 1

$$\int\frac{1}{\sqrt{9-4x^2}}\,\mathrm{d}x=\frac 13\int\frac{1}{\sqrt{1-\frac 49x^2}}\,\mathrm{d}x$$ Then substitute $t=\frac 23 x \Rightarrow \mathrm{d}x=\frac 32 \mathrm{d}t$ to obtain $$\frac 12\int\frac{1}{\sqrt{1-t^2}}\,\mathrm{d}t $$ What is the derivative of $\arcsin t$?

Answer 2

HINT

After the substitution $x =\frac{3}{2}\sin u$ we get $$ \sqrt{9-4x^2} = \sqrt{9 - 9\sin^2 u} = 3 \cos u. $$

Can you complete this?

Answer 3

Let $\sin t = \frac23 x$. Then,

$$\int\frac{dx}{\sqrt{9-4x^2}} =\frac12 \int \frac{\cos t dt}{\sqrt{1-\sin^2 t}} =\frac 12 \int dt =\frac12t+C=\frac12 \sin^{-1}\frac{2x}3+C$$

Answer 4

Let $x=\frac{3}{2}\sin\left(t\right)$ then $dx=\frac{3}{2}\cos\left(t\right)d\left(t\right)$:

$$\int_{2}^{4}\frac{1}{\sqrt{9-4x^{2}}}$$$$=\frac{1}{2}\int_{ }^{ }d\left(t\right)=\frac{1}{2}t+C$$$$=\frac{1}{2}\arcsin\left(\frac{2}{3}x\right)+C$$