Mathematics Asked by user801303 on November 24, 2020

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

**My attempt**:

I tried using substitution $x=sectheta$, $dx=sectheta tantheta dtheta$,

$$int (sec^2theta-1)(sec^3theta-3sectheta)^{4/3}

sectheta tantheta dtheta $$

$$=int tan^2theta sec^4theta(1-3cos^2theta)^{4/3} sectheta tantheta dtheta $$

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

$$=intdfrac{ sin^3theta}{ cos^8theta}(1-3cos^2theta)^{4/3} dtheta $$

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

Please help me solve this integral.

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).

Correct answer by user2661923 on November 24, 2020

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

$$int (x^2-1)(x^3-3x)^{4/3} mathop{dx}=int t^{4/3}frac{dt}{3}$$ $$=frac13frac{t^{7/3}}{7/3}+C$$$$=frac{(x^3-3x)^{7/3}}{7}+C$$

or alternatively,

$$int (x^2-1)(x^3-3x)^{4/3} dx=frac13int (3x^2-3)(x^3-3x)^{4/3} dx$$ $$=frac13int (x^3-3x)^{4/3} d(x^3-3x)$$ $$=frac13frac{(x^3-3x)^{7/3}}{7/3}+C$$ $$=frac{(x^3-3x)^{7/3}}{7}+C$$

Answered by Harish Chandra Rajpoot on November 24, 2020

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 $$ begin{split} int (x^2 -1)(x^3 - 3x)^{4/3}dx &= frac{1}{3}int (3x^2-3)(x^3-3x)^{4/3} dxcr &= frac{1}{3} int u^{4/3} du cr &= frac{1}{3} times frac{3u^{7/3}}{7} + C cr &= frac{1}{7} (x^3 - 3x)^{7/3} + C cr end{split} $$

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