# Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$

I need to solve the integral below, but I just can't figure how.

$$\int \sqrt{16-9x^2}\,dx$$

I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please explain to me how to finish it...

$$\frac {16}3 \int \cos^2\theta \,d\theta\$$

• find another expression for $\cos^2 (x)$ Jul 28, 2015 at 20:15
• you're on the right track Jul 28, 2015 at 20:21
• hint: $\cos 2\theta = \cos^2\theta - \sin^2\theta$ Jul 28, 2015 at 21:18

Hint: $$\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))$$

Edit: Also, after some calculations, I am pretty sure you made an error (can anyone double check this?)

You have got $\displaystyle\frac{4}{3}\int\cos^2\theta d\theta$, but the substitution $9x^2=16\sin^2\theta$ gives us the following:

$$\sqrt{16-9x^2}=\sqrt{16-16\sin^2\theta}=\sqrt{16(1-\sin^2\theta)}=\sqrt{16\cos^2\theta}=4\cos\theta$$

and since $x=\frac{4}{3}\sin\theta$, we have got $dx=\frac{4}{3}\cos\theta d\theta$, so

$$\displaystyle\int\sqrt{16-9x^2}dx=\int4\cos\theta\cdot\left(\frac{4}{3}\cos\theta d\theta\right)=\frac{16}{3}\int\cos^2\theta d\theta$$

and here is the way to do the rest

$\displaystyle\frac{16}{3}\int\cos^2\theta d\theta=\frac{16}{3}\int\left(\frac{1}{2}\left(1+\cos 2\theta)\right)\right)=\frac{8}{3}\int\left(1+\cos 2\theta\right)=\frac{8}{3}\left(\theta +\frac{1}{2}\sin2\theta+C\right)$. Since $\sin\theta =\frac{3x}{4}$, we have $\theta=\sin^{-1}\left(\frac{3x}{4}\right)$ and $\sin2\theta = \sqrt{16-9x^2}$ and so $\displaystyle\int\sqrt{16-9x^2}dx$ evaluates to $\displaystyle\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+\frac{1}{2}x\sqrt{16-9x^2}+C$

Edit: (To address OP's questions, as this is comment is not properly showing up in the comments section). After multiplying $\frac{8}{3}$ with $\frac{1}{2}\sin\theta$, we have $\displaystyle\frac{4}{3}\sin2\theta=\frac{4}{3}(2\sin\theta\cos\theta)=\frac{8}{3}\sin\theta\cos\theta=\frac{8}{3}\cdot\left(\frac{3x}{4}\right)\cdot\left(\frac{\sqrt{16-9x^2}}{4}\right)$, then cancel you get it.

• Thank you for noticing this. In fact, I did the right thing, don't know why I wrote 4/3.
– Mart
Jul 28, 2015 at 20:37
• Thanks again. but how do you that sin 2θ = $\sqrt{16-9x^2}$
– Mart
Jul 28, 2015 at 20:55
• @Mart I edited the post to address this, we use $\sin2\theta=2\sin\theta\cos\theta$
– user265675
Jul 28, 2015 at 21:17

Hint So far so good! The standard technique for handling this integrand is to invoke the double angle identity $$\cos 2 \theta = 2 \cos^2 \theta - 1.$$

$$\int\sqrt{16-9x^2}dx=$$

(Substitute $x=\frac{4\sin(u)}{3}$ and $dx=\frac{4\cos(u)}{3}du$ then $\sqrt{16-9x^2}=\sqrt{16-16\sin^2(u)}=4\cos(u)$ and $u=\sin^{-1}\left(\frac{3x}{4}\right)$):

$$\frac{4}{3}\int 4\cos^2(u)du=$$ $$\frac{16}{3}\int \cos^2(u)du=$$ $$\frac{16}{3}\int \left(\frac{1}{2}\cos(2u)+\frac{1}{2}\right)du=$$ $$\frac{8}{3}\int \cos(2u)du+\frac{8}{3}\int 1 du=$$

(Substitute $s=2u$ and $ds=2du$):

$$\frac{4}{3}\int \cos(s)ds+\frac{8}{3}\int 1 du=$$ $$\frac{4\sin(s)}{3}+\frac{8}{3}\int 1 du=$$ $$\frac{4\sin(s)}{3}+\frac{8u}{3}+C=$$ $$\frac{4\sin(2u)}{3}+\frac{8u}{3}+C=$$ $$\frac{8u}{3}+\frac{8}{3}\sin(u)\cos(u)+C=$$ $$\frac{8u}{3}+\frac{8}{3}\sin(u)\sqrt{1-\sin^2(u)}+C=$$ $$\frac{8\left(\sin^{-1}\left(\frac{3x}{4}\right)\right)}{3}+\frac{8}{3}\sin\left(\left(\sin^{-1}\left(\frac{3x}{4}\right)\right)\right)\sqrt{1-\sin^2\left(\left(\sin^{-1}\left(\frac{3x}{4}\right)\right)\right)}+C=$$

$$\frac{1}{2}\sqrt{16-9x^2}x+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+C$$

So:

$$\int\sqrt{16-9x^2}dx=\frac{1}{2}\sqrt{16-9x^2}x+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+C$$

$\frac{16}{3}\int(cos^2\theta)$

= $\frac{16}{3}\int\frac{1-cos2\theta}{2}$

= $\frac{16}{6}\int(1-cos2\theta)$

= $\frac{16}{6}(\int(1) - \int(cos2\theta))$

= $\frac{16}{6}\theta$ - $\frac{16}{6}\int(cos2\theta)$

= $\frac{8}{3}\theta$ - $\frac{8}{3}\int(cos2\theta)$

= $\frac{8}{3}\theta$ - $\frac{8}{3}(\frac{1}{2})\int(cos(v))$, substitute $v = 2\theta$

= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(v)$

= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(2\theta)$

= $\frac{8}{3}\theta$ + $\frac{4}{3}(2) sin(\theta) cos (\theta)$