Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.

I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody please check against my answer and see if I made a mistake.

  • 2
    How did you get this answer? – Michael Galuza Jul 27 '15 at 2:19
  • 1
    @Dr.MV Thanks for giving the detailed answer! This is a question from Trinity College Cambridge sample Interview test. – Rescy_ Jul 27 '15 at 4:15
  • If you showed your work, it would be easier to see where any mistakes might be. – robjohn Jul 27 '15 at 4:23
  • You're welcome. My pleasure. And hope that the answer helps you with the interview. Best wishes. ;-) – Mark Viola Jul 27 '15 at 4:23
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    you can always check your answer on wolframalpha.com – Mark Jul 27 '15 at 5:59
up vote 5 down vote accepted

Integrating by parts, we have

$$\begin{align} I_n&=\int_0^1\frac{x^n}{(1+x^3)^{1/2}}dx\\\\ &=\left.\frac23 x^{n-2}(x^3+1)\right|_0^1-\frac23(n-2)\int_0^1x^{n-3}(x^3+1)^{1/2}dx \tag 1\\\\ &=\frac232^{1/2}-\frac23(n-2)\int_0^1\frac{x^{n-3}(x^3+1)}{(x^3+1)^{1/2}}\,dx \\\\ &=\frac232^{1/2}-\frac23 (n-2)I_n-\frac23(n-2)I_{n-3}\\\\ (2n-1)I_n+2(n-2)I_{n-3}&=2^{3/2} \tag 2\\\\ \end{align}$$

Using $(1)$, we can see that $I_2$ given by

$$I_2=\frac23(\sqrt{2}-1) \tag 3$$

Then, using $(2)$ and $(3)$ and iterating once to $I_5$

$$I_5=\frac29 (2-\sqrt{2})$$

and again to $I_8$ reveals


  • Fun. When I started to post - no answers were given. When I finished you had the perfect answer ;) UP! – johannesvalks Jul 27 '15 at 3:07
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    @johannesvalks And yours! +1 – Mark Viola Jul 27 '15 at 3:10

Perform the integration in steps:

$$ \begin{array}{rcl} I_8 &=& \displaystyle \int_0^1 \frac{x^8}{\sqrt{x^3 + 1}} dx = \displaystyle \int_0^1 x^6 \frac{x^2}{\sqrt{x^3 + 1}} dx\\ &=& \displaystyle \left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1 - \int_0^1 4 x^3 x^2 \sqrt{x^3+1} dx\\ &=& \displaystyle \left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1 - \left[ \frac{8}{9} x^3 \sqrt{x^3+1}^3 \right]_0^1 + \int_0^1 \frac{8}{3} x^2 \sqrt{x^3+1}^3 dx\\ &=& \displaystyle \left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1 - \left[ \frac{8}{9} x^3 \sqrt{x^3+1}^3 \right]_0^1 + \left[ \frac{16}{45} \sqrt{x^3+1}^5 \right]_0^1\\ &=& \displaystyle \left[ \frac{2}{3} x^6 \sqrt{x^3+1} - \frac{8}{9} x^3 \sqrt{x^3+1}^3 + \frac{16}{45} \sqrt{x^3+1}^5 \right]_0^1\\ &=& \displaystyle \left( \frac{2}{3} - \frac{16}{9} + \frac{64}{45} \right) \sqrt{2} - \frac{16}{45}\\ &=& \displaystyle \bbox[16px,border:2px solid #800000] {\frac{14}{45} \sqrt{2} - \frac{16}{45}} \end{array} $$

\begin{align*}I_n&=\int_0^1 \dfrac{x^n}{\sqrt{x^3+1}}dx=\int_0^1 \dfrac{x^{n-3}(x^3+1-1)}{\sqrt{x^3+1}}dx = \int_0^1 x^{n-3}\sqrt{x^3+1}dx - \int_0^1 \dfrac{x^{n-3}}{\sqrt{x^3+1}}dx\\ &= \int_0^1 x^{n-3}\sqrt{x^3+1}dx-I_{n-3}\end{align*}This integral is handled with integration by parts: $$\int_0^1 x^{n-3}\sqrt{x^3+1}dx=\frac{x^{n-2}}{n-2}\sqrt{x^3+1}|_0^1-\frac{3}{2(n-2)}\int_0^1 \dfrac{x^{n}}{\sqrt{x^3+1}}dx=\dfrac{\sqrt{2}}{n-2}-\frac{3}{2(n-2)}I_n$$Therefore $$I_n=\dfrac{\sqrt{2}}{n-2}-\frac{3}{2(n-2)}I_n-I_{n-3}$$ or $$(n-2+\frac{3}{2})I_n+(n-2)I_{n-3}=\sqrt{2}\tag{1}$$Which is equivalent to your formula, so I would guess you're correct, though I haven't computed $I_2$ to check if $I_8$ is correct.

If we set $$ I_n=\int_0^1\frac{x^n}{\sqrt{x^3+1}}\,\mathrm{d}x\tag{1} $$ Then, integration by parts gives $$ \begin{align} I_{n+3}+I_n &=\int_0^1\frac{x^n(x^3+1)}{\sqrt{x^3+1}}\,\mathrm{d}x\\ &=\int_0^1x^n\sqrt{x^3+1}\ \mathrm{d}x\\ &=\frac1{n+1}\int_0^1\sqrt{x^3+1}\ \mathrm{d}x^{n+1}\\ &=\frac1{n+1}\left[\sqrt2-\int_0^1\frac32\frac{x^{n+3}}{\sqrt{x^3+1}}\,\mathrm{d}x\right]\tag{2} \end{align} $$ Applying a bit of algebra to $(2)$ yields $$ \left(2n+5\right)I_{n+3}+(2n+2)I_n=2\sqrt2\tag{3} $$ which, after substituting $n\mapsto n-3$, gives $$ \bbox[5px,border:2px solid #C0A000]{(2n-1)I_n+(2n-4)I_{n-3}=2\sqrt2}\tag{4} $$ Next, substitute $x\mapsto(x-1)^{1/3}$: $$ \begin{align} I_8 &=\int_0^1\frac{x^8}{\sqrt{x^3+1}}\,\mathrm{d}x\\ &=\frac13\int_1^2\frac{(x-1)^2}{\sqrt{x}}\,\mathrm{d}x\\ &=\frac13\left[\frac25x^{5/2}-\frac43x^{3/2}+2x^{1/2}\right]_1^2\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{14\sqrt2-16}{45}}\tag{5} \end{align} $$

  • This is pretty much the same as some of the other answers, but since their values for $I_8$ and the one given in the question differ, I will leave this as confirmation of the ones in the other answers. – robjohn Jul 27 '15 at 4:21
  • I really like how I can see the style of an answer and know who it was without finishing. :D – Simply Beautiful Art Mar 1 '17 at 1:38

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