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...
$$ \frac {16}3 \int \cos^2\theta \,d\theta\ $$
On
Hint So far so good! The standard technique for handling this integrand is to invoke the double angle identity $$\cos 2 \theta = 2 \cos^2 \theta - 1.$$
On
$$\frac{16}{3}\int\cos^2\theta\,d\theta=\frac{16}{3}\int\frac{1+\cos2\theta}{2}d\,\theta=\frac{8}{3}\int(1+\cos2\theta)d\,\theta=\frac{8}{3}(\theta+\frac{1}{2}\sin2\theta)$$
$$9x^2=16\sin^2\theta\Rightarrow\sin\theta=\frac{3x}{4}\Rightarrow\theta=sin^{-1}(\frac{3x}{4})$$
$$\sin\theta=\frac{3x}{4}\Rightarrow \cos\theta=\sqrt{1-sin^2\theta}=\frac{\sqrt{16-9x^2}}{4}$$
$$\sin2\theta=2\sin\theta\cos\theta=\frac{3x\sqrt{16-9x^2}}{8}$$
so the answer is:
$$\int \sqrt{16-9x^2}\,dx=\frac{16}{3}\int\cos^2\theta\,d\theta=\frac{8}{3}(\theta+\frac{1}{2}\sin2\theta)=\frac{8}{3}(\sin^{-1}(\frac{3x}{4})+\frac{3x\sqrt{16-9x^2}}{16})$$
On
$$\int\sqrt{16-9x^2}dx=$$
(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:
$$\int\sqrt{16-9x^2}dx=\frac{1}{2}\sqrt{16-9x^2}x+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+C$$
On
$\frac{16}{3}\int(cos^2\theta)$
= $\frac{16}{3}\int\frac{1-cos2\theta}{2}$
= $\frac{16}{6}\int(1-cos2\theta)$
= $\frac{16}{6}(\int(1) - \int(cos2\theta))$
= $\frac{16}{6}\theta$ - $\frac{16}{6}\int(cos2\theta)$
= $\frac{8}{3}\theta$ - $\frac{8}{3}\int(cos2\theta)$
= $\frac{8}{3}\theta$ - $\frac{8}{3}(\frac{1}{2})\int(cos(v))$, substitute $v = 2\theta$
= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(v)$
= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(2\theta)$
= $\frac{8}{3}\theta$ + $\frac{4}{3}(2) sin(\theta) cos (\theta)$
= $\frac{8}{3}arcsin(\frac{3}{4}x)$ + $\frac{8}{3}(\frac{3x}{4})$$\frac{sqrt (16-9x^2)}{4}$
= $\frac{8}{3}arcsin(\frac{3}{4}x)$ + $\frac{1}{2}x[sqrt (16-9x^2)]$ + C
On
$$ \begin{eqnarray} \color{blue}{ \int \sqrt{ 16 - 9 x^2 } d x } &=& x \sqrt{ 16 - 9 x^2 } + \int \frac{9 x^2}{ \sqrt{ 16 - 9 x^2 } } dx\\ &=& x \sqrt{ 16 - 9 x^2 } - \int \frac{16 - 9 x^2}{ \sqrt{ 16 - 9 x^2 } } dx + \int \frac{16}{ \sqrt{ 16 - 9 x^2 } } dx\\ &=& x \sqrt{ 16 - 9 x^2 } - \underbrace{ \color{blue}{ \int \sqrt{ 16 - 9 x^2 } dx } }_{\displaystyle \text{This is the same!}} + \int \frac{16}{ \sqrt{ 16 - 9 x^2 } } dx. \end{eqnarray} $$
$$ \begin{eqnarray} \color{blue}{ \int \sqrt{ 16 - 9 x^2 } d x } &=& \frac{1}{2} x \sqrt{ 16 - 9 x^2 } + \int \frac{8}{ \sqrt{ 16 - 9 x^2 } } dx\\ &=& \frac{1}{2} x \sqrt{ 16 - 9 x^2 } + \frac{8}{3} \underbrace{ \color{green}{ \int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big) } }_{\displaystyle \text{This should be familiar!}^{(1)}}\\ \\ &=& \bbox[16px,border:2px solid #800000] { \color{#800000}{ \frac{1}{2} x \sqrt{ 16 - 9 x^2 } + \frac{8}{3} {\sin^{-1}}\big( 3 x / 4\big) } } \end{eqnarray} $$
(1) In case $$ \color{green}{ \int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big) } $$ is not familiar...
$$ \begin{eqnarray} \color{green}{ \int \frac{1}{ \sqrt{ 1 - \big( 3 x / 4 \big)^2 } } d\big( 3 x / 4 \big) } &\stackrel{\color{magenta}{3 x / 4 = \sin(\phi)}}=& \int \frac{1}{\sqrt{1 - \sin^2(\phi)} } d\sin(\phi)\\ &=& \int \frac{\cos(\phi)}{\cos(\phi)} d\phi\\ &=& \phi\\ &\stackrel{\color{magenta}{\phi = {\sin^{-1}}\big( 3 x / 4\big)}}=& {\sin^{-1}}\big( 3 x / 4\big). \end{eqnarray} $$
On
Notice, this standard formula $$\int\sqrt{a^2-x^2}dx=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\left(\frac{x}{a}\right)\right)$$ Hence, we have $$\int\sqrt{16-9x^2}dx=\int\sqrt{(4)^2-(3x)^2}dx$$ let $3x=u\implies 3dx=du$$$=\frac{1}{3}\int\sqrt{(4)^2-(u)^2}du $$ $$=\frac{1}{3}\cdot\frac{1}{2}\left(u\sqrt{(4)^2-u^2}+(4)^2\sin^{-1}\left(\frac{u}{4}\right)\right)$$ $$=\frac{1}{6}\left(3x\sqrt{(4)^2-(3x)^2}+16\sin^{-1}\left(\frac{3x}{4}\right)\right)$$ $$=\color{blue}{\frac{1}{2}x\sqrt{16-9x^2}+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)}$$
Hint: $$\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
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.