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Evaluate$$\int x\sqrt{x^2 - 4}\,dx$$using trigonometric functions.

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Hint: Substitution, not trig. –  André Nicolas Mar 17 '14 at 20:26
@ahmedsalah Are you familiar with the u-sub? –  imranfat Mar 17 '14 at 20:26
@ahmed salah, you accepted an answer that is a blatant copy of another user's answer. A pity...and unfair. –  DonAntonio Mar 19 '14 at 18:25

3 Answers 3

up vote 4 down vote accepted

To evaluate $$\int x\sqrt{x^2 - 4}\,dx$$

substitute $\quad u = x^2 - 4 \implies du = 2x\,dx \iff \dfrac 12\,du = x\,dx.$

This gives us the integral $$\frac 12 \int u^{1/2}\,du$$

This gives us the integral $$\begin{align}\int x\sqrt{x^2 - 4} \,dx & = \int (\underbrace{x^2 - 4}_{u})^{1/2}\,\underbrace{x\,dx}_{\frac 12 \,du}\\ \\ & = \frac 12 \int u^{1/2}\,du \\ \\ & =\frac 12 \dfrac {u^{3/2}}{3/2} +C \\ \\ & = \frac 13 u^{3/2} + C\end{align}$$

Now, we just need to "back substitute" $\,u = x^2 - 4\,$ to get our final answer $$\frac 13(x^2 - 4)^{3/2} + C$$

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But you didn't use trigonometric functions! –  GEdgar Mar 17 '14 at 21:56
A pity to be masochist and use trigonometric substitution in such an elementary case.+1 –  DonAntonio Mar 19 '14 at 18:24

You can interpret this as a trigonometric integration problem, but it leads in a big circle. With $x = 2 \sec \theta$, $dx = 2 \sec \theta \tan \theta \; d\theta$ and $\sqrt{x^2 - 4} = 2 \tan \theta$, so $$ \int x \sqrt{x^2 - 4} \; dx = \int (2 \sec \theta)(2 \tan \theta)(2 \sec \theta \tan \theta \; d\theta) = 8 \int \sec^2 \theta \tan^2 \theta \; d\theta. $$ But, in order to evaluate this integral you need to make a substitution, such as $u = \tan \theta$, so $du = \sec^2 \theta \; d\theta$. Now, $$ \begin{align} 8 \int \sec^2 \theta \tan^2 \theta \; d\theta &= 8 \int u^2 \; du \\ &= \frac{8}{3} u^3 + C \\ &= \frac{8}{3} \tan^3 \theta + C \\ &= \frac{8}{3} \left( \frac{(x^2 - 4)^{1/2}}{2} \right)^3 + C \\ &= \frac{1}{3} \left( x^2 - 4 \right)^{3/2} + C. \end{align} $$ Note that in hindsight, you can see that that $$ u = \tan \theta = \frac{\sqrt{x^2 - 4}}{2}, $$ which is essentially the substitution that you would make (probably without the factor of $2$) if you weren't trying to use trigonometric substitution.

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Another approach, perhaps simpler and definitely shorter:

$$\int x\sqrt{x^2-4}\,dx=\frac12\int(x^2-4)'\sqrt{x^2-4}\,dx=\frac12\frac23(x^2-4)^{3/2}+C=\ldots$$

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Oh, yes it is @amWhy. First, it is obviously much shorter and, imo, simpler. Second, as it is not a substitution we don't need to check differentials equivalencies, go back to the original variable, etc. In fact, many (perhaps most?) of the basic, usual integrals done by substitution can be done this way, leaving thus substitution for the really tough ones. –  DonAntonio Mar 17 '14 at 21:00
It's only shorter looking because I happened to go through more labor intensive steps (perhaps more than I should have, since I left little for the OP). After designating u, I could have done precisely what you did. Any way, I didn't mean to argue, seriously! I'll delete my comment, and for the sake of sportsmanship $\to (+1)$ –  amWhy Mar 17 '14 at 21:03
I didn't see the previous comments, but I can guess what happened. For what it's worth, in my undergrad. we always did these integrals like this, we were never even taught substitution here. The first time I saw such a thing was here on MSE, then I realised it's commonplace worldwide. –  Git Gud Mar 19 '14 at 16:50

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