3
$\begingroup$

I would appreciate your help solving this integral:

$$\int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x$$

I tried using linear substitution with $ t = x/a $ and then trying to bring it to some combination of the known integral of $\arctan (x) = \frac{1}{x^2+1}$ but I'm not sure it will be helpful because there isn't just $1 $ in the numerator.

Basically, I got stuck very early in the process:

$$\int_2^3 \frac{1+x^3}{(a^2(\frac{x^2}{a^2}+1))^\frac{3}{2}}\mathrm{d}x$$

Thank you.

$\endgroup$
  • 2
    $\begingroup$ Ah I see your confusion, pretty hard to compute an integral without a differential. $\endgroup$ – electronpusher May 6 '17 at 21:24
  • 1
    $\begingroup$ The antiderivative of $$\frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}$$ is $$\frac{2a^4+a^2x^2+x}{a^2\sqrt{a^2+x^2}}+C.$$ $\endgroup$ – zoli May 6 '17 at 21:38
  • $\begingroup$ @zoli, I got $frac{x+a^2\sqrt{x^2+a^}+2a^4}{a^2\sqrt{x^2+a^2}}$ $\endgroup$ – electronpusher May 6 '17 at 21:43
  • 1
    $\begingroup$ wolframalpha.com/input/… $\endgroup$ – zoli May 6 '17 at 21:45
1
$\begingroup$

I suggest breaking it into two integrals, and using the trig substitution $x=a\tan t$ to turn each one into a trigonometric integral. Thus:

$$\begin{align} \int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x &= \int_2^3 \frac{dx}{(x^2+a^2)^{3/2}} + \int_2^3\frac{x^3}{(x^2+a^2)^{3/2}}dx\\ &=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sec^2 t}{a^3\sec^3 t}dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a^3 \tan^3 t \cdot a\sec^2 t}{a^3\sec^3 t}dt\\ &=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{1}{a^2}\cos t dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sin^3 t}{\cos^2 t}dt \end{align}$$

Can you get it from there?

$\endgroup$
  • $\begingroup$ Thank you for commenting but no so sure I followed - what is the meaning of $\sec$? $\endgroup$ – Noam May 6 '17 at 22:29
  • 1
    $\begingroup$ $\sec$ is the trigonometric function secant, the reciprocal of cosine. $\endgroup$ – G Tony Jacobs May 7 '17 at 0:02
1
$\begingroup$

Hint: Let $x \mapsto a \tan \theta$, then $\mathrm{d}x = a(1 + \tan \theta) \,\mathrm{d} \theta$. Then

$$ \int \frac{1+x^3}{(a^2+x^2)^{3/2}} = \frac{1}{a^2}\int \frac{\sin^3\theta}{\cos ^2\theta}+\cos \theta\,\mathrm{d}\theta $$

For the first integral use $\sin^3 \theta = \sin \theta(1-\cos^2\theta)$ and set $y \mapsto \cos \theta$. The latter integral is trivial.

$\endgroup$
1
$\begingroup$

Your integral can be split into $$\int\frac{dx}{(x^2+a^2)^{3/2}}+\int\frac{x^2\cdot xdx}{(x^2+a^2)^{3/2}}$$

The first integral can be solved by trig substitution. U-substitution can be used on the second integral to obtain

$$\int\frac{x^2\cdot xdx}{(x^2+a^2)^{3/2}} =\frac{1}{2}\int\frac{u-a^2}{u^{3/2}}du \\ = \frac{1}{2}\int(u^{-1/2}-a^2u^{-3/2})du$$.

And it's busywork from there.

$\endgroup$
  • $\begingroup$ thank you for commenting but I think you used $x^2$ instead of $x^3$? and can you elaborate on the trig substitution you made? $\endgroup$ – Noam May 6 '17 at 22:31
  • $\begingroup$ I used the same trig sub explained in G Tony Jacobs' answer; however, I did not apply it to the second term. For the second term I used the substitution $$u=x^2+a^2$$. I separated $$x^3=x^2 \cdot x$$ to obtain $$xdx=1/2du$$, and the leftover $$x^2=u-a^2$$ (according to the substitution I defined). I recommend trying to go through the steps yourself to fully understand. $\endgroup$ – electronpusher May 7 '17 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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