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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.

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

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

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

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  • $\begingroup$ 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
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    $\begingroup$ $\sin^2 u \cos^2 u = \sin^2 u(1 - \sin^2 u) = \sin^2 u - \sin^4$. $\endgroup$ – 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
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You also may use the method Differential Binomial to use another substitution $$a^2-x^2=x^2t^2$$ Try it. It works good. :-)

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$$ \int x^2 \sqrt{a^2 - x^2} ~dx ~ = ~ a^4 \int \sin^2 u \cos^2 u ~du ~ = ~ a^4 \int \sin^2 u - \sin^4 u ~ du \quad (x = a \sin u)$$ 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$$a^4 \int \left(\sin^2 u - \sin^4 u \right) ~ du =\frac{a^4}{2} \int \left(1-\cos{2u} \right) \ du \ -\frac{a^4}{8} \int \left( 3-4\cos 2u+\cos{4u} \right) \ du$$

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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 $J$.

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