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$$\frac{\int_{0}^{a} x^4 \sqrt{a^2 - x^2} \ dx}{\int_{0}^{a} x^2 \sqrt{a^2 - x^2} \ dx}$$

I tried solving this using trigonometric substitution, and it transforms into a very lengthy expression. Is there any easier way to do this problem?

EDIT After substitution the integral looks like this: $$= \frac{a^{6} \int \cos^{2}\left(u\right) \sin^{4}\left(u\right) \, du}{ a^{4} \int \cos^{2}\left(u\right) \sin^{2}\left(u\right) \, du}$$

( Substituting $x = a \sin (u)$ and $dx = a \cos (u) \, du$ in both numerator and denominator )

The denominator term looks relatively easy; however, the integral in the numerator is quiet lengthy.

rohit1729
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  • Do you want a solution for solving the integral in the numerator or just another method? – Bowei Tang Jul 31 '24 at 15:11
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    @rohit1729 The most efficient solution would be to use the beta function and avoid needing the antiderivatives altogether. – David H Jul 31 '24 at 15:18
  • If you define $I(n)=\int^{a}_{0} x^{2n}\sqrt{a^2-x^2}\ dx$, then Integration by parts gives $I(n)$ in terms of $I(n-1)$ and $I(0)$ is a quarter of the area of a circle. Not much quicker but another way. – RJG Jul 31 '24 at 15:24
  • After your trigonometric substitution, use the change of variables : $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ – PandaScientist Jul 31 '24 at 16:07

5 Answers5

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Let $$ K=\int_0^{\frac\pi2} \cos^{2}\left(u\right) \sin^{2}\left(u\right)\,du. $$ Noting \begin{align} L&:=\int_0^{\frac\pi2} \cos^{2}\left(u\right) \sin^{4}\left(u\right) \, du=\int_0^{\frac\pi2} \cos^{2}\left(u\right) \sin^{2}\left(u\right)(1-\cos^2(u)) \, du \\ &=K-\int_0^{\frac\pi2}\cos^4(u)\sin^2(u) \, du=K-L \end{align} one has $$ L=\frac12K $$ and hence $$\frac{\int_{0}^{a} x^4 \sqrt{a^2 - x^2} \ dx}{\int_{0}^{a} x^2 \sqrt{a^2 - x^2} \ dx}=\frac{a^2}{2}.$$

Sine of the Time
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xpaul
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Alternatively, integrate by parts as follows

\begin{align}\int_{0}^{a} x^4 \sqrt{a^2 - x^2} \ dx =&\int_{0}^{a} \frac{x^3}{(a^2-x^2)^{3/2}}\ d\bigg[ -\frac16(a^2-x^3)^3\bigg]\\ \overset{ibp}=& \ \frac{a^2}2 \int_{0}^{a} x^2 \sqrt{a^2 - x^2} \ dx \end{align}

Quanto
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4

In the numerator, introducing the limit, we get: $$ \int_0^{\frac{\pi}{2}}\cos^2u\sin^4u du=B\left(\frac52,\frac32\right)=\frac 12 \frac{\Gamma(\frac52)\Gamma(\frac32)}{\Gamma(4)} $$ where $\Gamma$ is the gamma function, in case you do not know. It has a value of: $$ B\left(\frac52,\frac32\right)=\frac{\pi}{32} $$ Similarly, the denominator is: $$ \int_0^{\frac{\pi}{2}}\cos^2u\sin^2u du=B\left(\frac32,\frac32\right)=\frac {\pi}{16} $$ Therefore, the value of the total integral is: $$\frac{a^2}{2}$$

M.Riyan
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As defined in my comment above,

$I(n) = \int^a_0 x^{2n-1}\ x\sqrt{a^2-x^2}\ dx$ which by parts is

$[-\frac{1}{3}x^{2n-1}(a^2-x^2)^{\frac{3}{2}}]^a_0+\frac{1}{3}\int^a_0 (2n-1)x^{2n-2}(a^2-x^2)^{\frac{3}{2}}\ dx $

which is $\frac{1}{3}(2n-1)(a^2 I(n-1)-I(n))$

Thus $I(n)=\dfrac{2n-1}{2n+2}a^2 I(n-1).$

Thus you want $\dfrac{I(2)}{I(1)}=\dfrac{3a^2}{6}=\dfrac{a^2}{2}$.

RJG
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The Wallis integrals $$W_n:=\int_0^{\pi/2}\sin^nu\,du$$ are known to satisfy $$nW_n=(n-1)W_{n-2}.$$ Therefore, $$\begin{align}\frac{\int_0^ax^n\sqrt{a^2-x^2}\,dx}{\int_0^ax^{n-2}\sqrt{a^2-x^2}\,dx}&=\frac{a^{n+2}\int\cos^2u\sin^nu\,du}{a^n\int\cos^2u\sin^{n-2}u\,du}\\&=a^2\frac{W_n-W_{n+2}}{W_{n-2}-W_n}\\&=a^2\frac{W_n-\frac{n+1}{n+2}W_n}{\frac n{n-1}W_n-W_n}\\&=a^2\frac{n-1}{n+2}.\end{align}$$

Anne Bauval
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