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How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$

My attempt:

I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,

$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3} \sec\theta\ \tan\theta d\theta $$

$$=\int \tan^2\theta \sec^4\theta(1-3\cos^2\theta)^{4/3} \sec\theta\ \tan\theta d\theta $$ $$=\int \tan^3\theta \sec^5\theta(1-3\cos^2\theta)^{4/3}\ d\theta $$

$$=\int\dfrac{ \sin^3\theta}{ \cos^8\theta}(1-3\cos^2\theta)^{4/3}\ d\theta $$

I can't see if this substitution will work or not. This has become so complicated.

Please help me solve this integral.

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    $\begingroup$ I upvoted to reverse someone's downvote. The OP clearly made a serious effort to solve the problem and included the details of his work in the query. It is not the OP's fault that he doesn't know which substitution to use. $\endgroup$ – user2661923 Jul 29 at 11:52
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I agree with the other answers. My response is long-winded so...

Often when attacking indefinite integrals, you will immediately suspect that a substitution [i.e. $u = g(x)$] is needed, but won't be sure which substitution to try.

I have to ask the OP:
Why did you think that $x = \sec \theta$ was the right substitution? Had you recently been exposed to problems that seemed similar where $x = \sec \theta$ was the right substitution?

The point of my response/rant is to develop the OP's intuition. Since the integral contains $(x^3 - 3x)^{(4/3)},$ my first guess as to the right substitution to try would be $u = (x^3 - 3x).$ This would convert this portion of the integral to $u^{(4/3)}.$

The idea is that (as a first guess for the right substitution), I would be hoping that (except for the $u^{(4/3)}$ factor), the remainder of the integral would be a polynomial in $u$, where each term has an integer exponent.

As I say, the point of my response is simply to expand the OP's intuition (and perspective).

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    $\begingroup$ Maybe the OP tried to substitute $\sec \theta$ because $\sec^2 \theta- 1 = \tan^2 \theta$. $\endgroup$ – Toby Mak Jul 30 at 0:39
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    $\begingroup$ But the difficulty was not the x^2- 1 part, It was the 3/4 power part. $\endgroup$ – user247327 Jul 30 at 0:41
  • $\begingroup$ @AlexanderGruber I've given your editing of my answer some thought and I've decided that your underlying point (against editorializing on mathSE) is well taken, and I will keep this in mind in the future. That being said, it does leave a gap that I see no way of filling: I think that it is important for the OP to know why he went wrong. Unfortunately, I see know way of clearly communicating that without editorializing. Consequently, I see no choice but (for my future mathSE responses) to ignore why the OP went wrong, still addressing (where appropriate) where the OP went wrong. $\endgroup$ – user2661923 Aug 2 at 0:37
  • $\begingroup$ @user2661923 Well, I think you did tell OP where they went wrong: they used a trig sub, which it turns out is not the correct approach for this integral. This is information we know, as it was included in OP's work. But the idea that they thought to use a trig sub because he has a bad teacher is just speculation. I support your idea of developing OP's intuition; however, I don't think the omitted paragraph actually serves that purpose. $\endgroup$ – Alexander Gruber Aug 2 at 2:09
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If you multiply and divide by $3$, you get $$ \int (x^2 -1)(x^3 - 3x)^{4/3}dx = \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx $$ changing variable to $u = x^3 - 3x$ you have $du = (3x^2 - 3x)dx$ so $$ \begin{split} \int (x^2 -1)(x^3 - 3x)^{4/3}dx &= \frac{1}{3}\int (3x^2-3)(x^3-3x)^{4/3} dx\cr &= \frac{1}{3} \int u^{4/3} du \cr &= \frac{1}{3} \times \frac{3u^{7/3}}{7} + C \cr &= \frac{1}{7} (x^3 - 3x)^{7/3} + C \cr \end{split} $$

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Let $x^3-3x=t\implies (3x^2-3)dx=dt$ or $(x^2-1)dx=\frac{dt}{3}$

$$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\int t^{4/3}\frac{dt}{3}$$ $$=\frac13\frac{t^{7/3}}{7/3}+C$$$$=\frac{(x^3-3x)^{7/3}}{7}+C$$

or alternatively,

$$\int (x^2-1)(x^3-3x)^{4/3}\ dx=\frac13\int (3x^2-3)(x^3-3x)^{4/3}\ dx$$ $$=\frac13\int (x^3-3x)^{4/3}\ d(x^3-3x)$$ $$=\frac13\frac{(x^3-3x)^{7/3}}{7/3}+C$$ $$=\frac{(x^3-3x)^{7/3}}{7}+C$$

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