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
  • 2
    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

$$I_8=\frac{2}{45}(7\sqrt{2}-8)$$

  • 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
  • 1
    @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

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.