Evalute This Integral Function Using Trigonometric Function?

Question

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

Answer 1

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 $$\frac 13(x^2 - 4)^{3/2} + C$$

Answer 2

Another approach, perhaps simpler and definitely shorter:

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

Answer 3

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.