Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 3 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 $$\frac 13(x^2 - 4)^{3/2} + C$$

share|cite|improve this answer
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.

share|cite|improve this answer

Another approach, perhaps simpler and definitely shorter:

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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