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How to evaluate the integral $$\int_0^1 (1-x^{2/3})^{3/2} \, dx.$$ Does some substitution work?

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u shud give a dx... –  soumajit das May 9 at 6:22
    
u shud add a dx..otherwise its invalid...@ ajinkya –  soumajit das May 9 at 6:24
    
The substitution $u=x^{2/3}$ would be an obvious first guess. –  David H May 9 at 6:24
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@soumajitdas Not necessarily. Some authors dispense with the dx notation altogether. –  David H May 9 at 6:28

4 Answers 4

up vote 3 down vote accepted

Hint: substitute $u=x^{1/3}$. This puts the integral in a form amenable to a simple trigonometric substitution.

Substituting $u=x^{1/3}$,

$$I=\int_{0}^{1}(1-x^{2/3})^{3/2}dx=3\int_{0}^{1}u^2\,(1-u^2)^{3/2}\,du.$$

Substituting $u=\sin{\theta}$,

$$I=3\int_{0}^{1}u^2\,(1-u^2)^{3/2}\,du=3\int_{0}^{\pi/2}\sin^2{\theta}\,(1-\sin^2{\theta})^{3/2}\cos{\theta}\,d\theta\\ =3\int_{0}^{\pi/2}\sin^2{\theta}\,\cos^4{\theta}\,d\theta\\ =3\int_{0}^{\pi/2}\cos^4{\theta}\,d\theta-3\int_{0}^{\pi/2}\cos^6{\theta}\,d\theta\\ =\frac{9\pi}{16}-\frac{15\pi}{32}\\ =\frac{3\pi}{32}.$$

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$$ \begin{align} \int_0^1(1-x^{2/3})^{3/2}\,\mathrm{d}x &=\int_0^1(1-u)^{3/2}\,\mathrm{d}u^{3/2}\\ &=\frac32\int_0^1(1-u)^{3/2}u^{1/2}\,\mathrm{d}u\\ &=\frac32\mathrm{B}\left(\frac52,\frac32\right)\\ &=\frac32\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)}{\Gamma(4)}\\ &=\frac32\frac{\frac32\frac12\Gamma\left(\frac12\right)\frac12\Gamma\left(\frac12\right)}{\Gamma(4)}\\ &=\frac{3\pi}{32} \end{align} $$


Alternatively, we can use $$ \begin{align} \int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x &=\frac14\int_0^{2\pi}\cos^{2n}(x)\,\mathrm{d}x\\ &=\frac14\int_0^{2\pi}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{2n}\,\mathrm{d}x\\ &=\frac14\int_0^{2\pi}4^{-n}\sum_{k=0}^{2n}\binom{2n}{k}e^{i(2n-2k)x}\,\mathrm{d}x\\ &=\frac\pi2\binom{2n}{n}4^{-n} \end{align} $$ and $$ \begin{align} \int_0^1(1-x^{2/3})^{3/2}\,\mathrm{d}x &=\int_0^1(1-u)^{3/2}\,\mathrm{d}u^{3/2}\\ &=\int_0^1(1-v^2)^{3/2}\,\mathrm{d}v^3\\ &=3\int_0^1(1-v^2)^{3/2}v^2\,\mathrm{d}v\\ &=3\int_0^{\pi/2}\cos^4(\theta)\sin^2(\theta)\,\mathrm{d}\theta\\ &=3\int_0^{\pi/2}\left[\cos^4(\theta)-\cos^6(\theta)\right]\,\mathrm{d}\theta\\ &=\frac{3\pi}{2}\left[\binom{4}{2}4^{-2}-\binom{6}{3}4^{-3}\right]\\ &=\frac{3\pi}{32} \end{align} $$

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$$\int_0^1\Big(1-\sqrt[n]x\Big)^m~=~\int_0^1\Big(1-\sqrt[m]x\Big)^n~=~\frac{m!\cdot n!}{(m+n)!}~=~{m+n\choose n}^{-1}~=~{m+n\choose m}^{-1}$$


$\qquad\quad$ In this case, $m=n=\dfrac32$ , with $\Gamma\bigg(\dfrac12\bigg)=\sqrt\pi,~$ and $n!=\Gamma(n+1)=n\cdot\Gamma(n)$.

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You can also use $x=\sin^3\theta$ and $dx=3\sin^2\theta\cos\theta\ d\theta$, then $$ \begin{align} \int_0^1\left(1-x^{\large\frac23}\right)^{\large\frac32}\ dx&=\int_0^{\large\frac\pi2}\left(1-\sin^2\theta\right)^{\large\frac32}\cdot 3\sin^2\theta\ \cos\theta\ d\theta\\ &=3\int_0^{\large\frac\pi2}\cos^4\theta(1-\cos^2\theta)\ d\theta\\ &=3\int_0^{\large\frac\pi2}(\cos^4\theta-\cos^6\theta)\ d\theta. \end{align} $$ The last integral can be solved by using integration by reduction formula, we will obtain $$ \begin{align} \int_0^1\left(1-x^{\large\frac23}\right)^{\large\frac32}\ dx&=3\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta-3\int_0^{\large\frac\pi2}\cos^6\theta\ d\theta\\ &=3\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta-3\left(\left.\frac16\cos^5\theta\sin\theta\right|_0^{\large\frac\pi2}+\frac56\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta\right)\\ &=\frac12\int_0^{\large\frac\pi2}\cos^4\theta\ d\theta. \end{align} $$ The rest I leave it to you.

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