2
$\begingroup$

I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$

and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$

I've tried using the substitution $x=a\sin t$ and $x=a\sec t$ Both have looked promising but I just can't seem to finish it out.

Thanks in advance for any help.

$\endgroup$
  • 4
    $\begingroup$ Using $x=a\sin t$ looks ok. $\endgroup$ – freak_warrior Nov 21 '13 at 13:05
7
$\begingroup$

We will assume that $a>0$ for the following.

We have $$\int x^2\sqrt{a^2-x^2} \,dx.$$

Let $$ \begin{align*} x&=a\sin{t} \\ dx &= a\cos{t} \, dt \\ a^2-x^2&=a^2-a^2\sin{t}\\ &=a^2\left( 1-\sin^2{t} \right)\\ &=a^2\cos^2{t}. \end{align*} $$

We substitute and integrate,

$$ \begin{align*} &\int a^2\sin^2t \cdot \sqrt{a^2\cos^2t}\,\cdot a \cos t \,dt \\ =&a^4\int\sin^2 t \cos^2 t \, dt \\ =&a^4 \int\left( \sin t \cdot \cos t \right)^2 \, dt \\ =&a^4 \int\left( \frac{1}{2}\sin(2t) \right)^2 \, dt \\ =&\frac{a^4}{4}\int \sin^2(2t)\, dt \\ =& \frac{a^4}{4} \int \frac{1-\cos(4t)}{2} \, dt \\ =& \frac{a^4}{8} \int \left(1-\cos(4t)\right) \, dt \\ =& \frac{a^4}{8} \left( t-\frac{1}{4}\sin(4t) \right)+c. \end{align*} $$ The back substitution will be simpler if we have single angled trig solutions, and so we can reduce, $$ \begin{align*} \sin(4t) &= 2\sin(2t)\cos(2t) \\ &=2\left( 2\sin t \cdot \cos t \left( \cos^2 t- \sin^2 t \right)\right) \\ &= 4\sin t \cos^3 t-4\sin^3 t \cos t. \end{align*} $$ Hence our integral is

$$\frac{a^4}{8}\left( t- \sin t \cos^3 t + \sin^3 t \cos t \right)+c.$$

For the back substitution, we have that $$x=a\sin t,$$ and so $$t=\sin^{-1}\left(\frac{x}{a}\right).$$

For the remaining part, we draw a right triangle with angle $t$, opposite side $x$, hypotenuse $a$, and it follows that the adjacent side will be $\sqrt{a^2-x^2}$.

We use the definition of $t$ and read straight from the right triangle to back substitute,

\begin{align*} & \frac{a^4}{8}\left( t- \sin t \cos^3 t + \sin^3 t \cos t \right) +c \\ =& \frac{a^4}{8}\left( \sin^{-1}\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)\left( \frac{\sqrt{a^2-x^2}}{a} \right)^3 +\left( \frac{x}{a} \right)^3\frac{\sqrt{a^2-x^2}}{a} \right) +c \\ =&\frac{a^4}{8}\left( \sin^{-1}\left(\frac{x}{a}\right) -\left(\frac{x}{a}\right)\frac{\sqrt{a^2-x^2}}{a}\left( \frac{a^2-x^2}{a^2}-\frac{x^2}{a^2} \right) \right) +c \\ =& \frac{a^4}{8}\left( \sin^{-1}\left(\frac{x}{a}\right)-\frac{x\sqrt{a^2-x^2}}{a^2}\left( \frac{a^2-2x^2}{a^2} \right) \right) +c \\ =&\frac{x}{8}\left( 2x^2-a^2 \right)\sqrt{a^2-x^2}+\frac{a^4}{8}\sin^{-1}\left(\frac{x}{a}\right)+c. \end{align*} This is the desired form.

$\endgroup$
  • $\begingroup$ @JWPerry, we can not write $\sqrt{a^2\cos^2t}=a\cos t$. In fact, $\sqrt{a^2\cos^2t}=|a\cos t|=|a|\cos t$, the reason has been explained in my answer $\endgroup$ – lab bhattacharjee Nov 22 '13 at 3:22
  • $\begingroup$ @labbhattacharjee + c. We are not integrating over the constant $a$. The difference is absorbed by $c$. The integral is verifiably correct. $\endgroup$ – J. W. Perry Nov 22 '13 at 4:03
  • $\begingroup$ @JWPerry, you are implicitly assuming that $a>0$ when you say hypotenuse $=a$ $\endgroup$ – lab bhattacharjee Nov 22 '13 at 4:06
  • $\begingroup$ @labbhattacharjee Tell me what you would modify here. Note this is a textbook integration and solution, but I appreciate your input. $\endgroup$ – J. W. Perry Nov 22 '13 at 4:11
  • $\begingroup$ @JWPerry, including the assumption explicitly $\endgroup$ – lab bhattacharjee Nov 22 '13 at 4:12
3
$\begingroup$

Putting $\displaystyle x=a\sin t,t=\arcsin\frac xa \implies -\frac\pi2\le t\le \frac\pi2$ as the principal value of inverse sine ratio lies $\displaystyle\left[-\frac\pi2,\frac\pi2\right]$

$\displaystyle\implies \cos t\ge0$

$\displaystyle\implies\sqrt{a^2-x^2}=\sqrt{a^2\cos^2t}=|a\cos t|=|a|\cos t$

$\displaystyle \int x^2\sqrt{a^2-x^2}dx=a^3|a|\int\sin^2t\cos^2tdt$

$$\text{Now as }\sin t\cos t=\frac{\sin2t}2, \sin^2t\cos^2t=\frac{\sin^22t}4$$

$$\text{Again as }\cos2u=\cos^2u-\sin^2u=1-2\sin^2u,\sin^22t=\frac{1-\cos4t}2$$

$$\int\sin^2t\cos^2tdt=\frac18\int(1-\cos4t)dt=\frac t8-\frac{\sin4t}{32}+C$$

Now as $\displaystyle x=a\sin t$ and $\sin4t=2\sin2t\cos2t=4\sin t\cos t\cos2t$

$\displaystyle\cos2t=1-2\sin^2t=1-2\left(\frac xa\right)^2=-\frac{2x^2-a^2}{a^2} $

As $\displaystyle\cos t\ge 0,\cos t=+\sqrt{1-\left(\frac xa\right)^2}=\frac{\sqrt{a^2-x^2}}{|a|}$

$\endgroup$
  • $\begingroup$ This does more than add something useful to the post, in particular the first 4 lines are enlightening. Thanks for setting me straight. $\endgroup$ – J. W. Perry Nov 22 '13 at 5:26
  • $\begingroup$ @J.W.Perry, my pleasure. We can do the same using $x=a\cos t$ as well based on the principal value $\endgroup$ – lab bhattacharjee Nov 22 '13 at 6:05
1
$\begingroup$

$$ \int x^2 \sqrt{a^2 - x^2} ~dx ~ = ~ a^4 \int \sin^2 u \cos^2 u ~du ~ = ~ a^4 \int \sin^2 u - \sin^4 u ~ du \quad (x = a \sin u)$$

$\endgroup$
  • $\begingroup$ Thanks, but I'm unsure of how you went from $sin^2ucos^2u$ to $sin^2u - sin^4u$? $\endgroup$ – Crockett Nov 21 '13 at 13:28
  • 1
    $\begingroup$ $\sin^2 u \cos^2 u = \sin^2 u(1 - \sin^2 u) = \sin^2 u - \sin^4$. $\endgroup$ – portin.daniel Nov 21 '13 at 13:31
  • $\begingroup$ Yes, I see it now. Thanks very much. $\endgroup$ – Crockett Nov 21 '13 at 13:32
  • $\begingroup$ Slightly simpler: $\sin ^2 u \cos^2 u = (0.5 \sin 2u)^2$ $\endgroup$ – Spine Feast Nov 21 '13 at 15:15
1
$\begingroup$

You also may use the method Differential Binomial to use another substitution $$a^2-x^2=x^2t^2$$ Try it. It works good. :-)

$\endgroup$
0
$\begingroup$

$$ \int x^2 \sqrt{a^2 - x^2} ~dx ~ = ~ a^4 \int \sin^2 u \cos^2 u ~du ~ = ~ a^4 \int \sin^2 u - \sin^4 u ~ du \quad (x = a \sin u)$$ Using the double angle formulae $\cos 2A=2\cos^2A-1=1-2\sin^2A$ $$\begin{align} \sin^2u &=\dfrac{1-\cos{2u}}{2}\\ \\ \sin^4u &=\left(\sin^2u\right)^2\\ \\ &=\left(\dfrac{1-\cos{2u}}{2} \right)^2\\ \\ &=\dfrac{1}{4} \left( 1-2\cos 2u+\cos^2 2u \right)\\ \\ &=\dfrac{1}{4} \left( 1-2\cos 2u+\dfrac{1+\cos{4u}}{2} \right)\\ \\ &=\dfrac{1}{8} \left( 3-4\cos 2u+\cos{4u} \right)\\ \end{align}$$ Hence$$a^4 \int \left(\sin^2 u - \sin^4 u \right) ~ du =\frac{a^4}{2} \int \left(1-\cos{2u} \right) \ du \ -\frac{a^4}{8} \int \left( 3-4\cos 2u+\cos{4u} \right) \ du$$

$\endgroup$
0
$\begingroup$

Calculating this integral can be done by parts: $$I=\int x^2\sqrt{a^2-x^2} dx= -\int x(\frac{1}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}} )' dx=$$ $$=-\frac{x}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}}+\frac{1}{3}\int(a^2 -x^2)\sqrt{a^2-x^2}dx. $$ $$\frac{4}{3}I=-\frac{x}{3}(\sqrt{a^2-x^2})^{\frac{3}{2}}+a^2\int\sqrt{a^2-x^2} dx. (1)$$ And$$J=\int\sqrt{a^2-x^2} dx =\int(x)'\sqrt{a^2-x^2} dx =x\sqrt{a^2-x^2} -\int\frac{a^2-x^2-a^2}{\sqrt{a^2-x^2}}dx=$$ $$=x\sqrt{a^2-x^2}-J+a^2\arcsin\frac{x}{a}$$ from which is obtained$$J=\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\arcsin\frac{x}{a}.$$ Substituting formula $I$ in $(1)$ we obtain $J$.

$\endgroup$

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.