How do you evaluate the integral #int x/(root3(x^2-1))#?

Question

Evelyn Agard · Accepted Answer

The integral equals #3/4(x^2 - 1)^(2/3) + C#

We will use u-substitution for this integral. Let #u = x^2 - 1#.

Then #du = 2xdx# and #dx= (du)/(2x)#. Call the integral #I#.

#I = int x/root(3)(u) * (du)/(2x)#

#I = 1/2 int 1/root(3)(u)#

#I = 1/2int u^(-1/3)#

#I = 1/2(3/2u^(2/3)) +C#

#I = 3/4u^(2/3) + C#

#I = 3/4(x^2 - 1)^(2/3) + C#

Hopefully this helps!

Luke Alvarez · Answer

undefined To evaluate the integral $\int \frac{x}{\sqrt{3(x^2-1)}} \, dx$, you can use a trigonometric substitution. Let $x = \sec(\theta)$, then $dx = \sec(\theta)\tan(\theta) \, d\theta$. Substituting these into the integral:

$$\int \frac{x}{\sqrt{3(x^2-1)}} \, dx = \int \frac{\sec(\theta)}{\sqrt{3(\sec^2(\theta)-1)}} \sec(\theta)\tan(\theta) \, d\theta$$

Simplify the expression inside the square root:

$$= \int \frac{\sec(\theta)}{\sqrt{3(\tan^2(\theta))}} \sec(\theta)\tan(\theta) \, d\theta$$

$$= \int \frac{\sec(\theta)}{\sqrt{3\tan^2(\theta)}} \sec(\theta)\tan(\theta) \, d\theta$$

$$= \int \frac{\sec^2(\theta)}{\sqrt{3}\tan(\theta)} \, d\theta$$

Now, recall that $\sec^2(\theta) - 1 = \tan^2(\theta)$. So, $\sqrt{3}\tan(\theta) = \sqrt{\sec^2(\theta) - 1}$.

Therefore,

$$= \int \frac{\sec^2(\theta)}{\sqrt{\sec^2(\theta) - 1}} \, d\theta$$

This integral can be simplified using a trigonometric identity. Let $u = \sec(\theta) - \tan(\theta)$, then $du = (\sec(\theta)\tan(\theta) + \sec^2(\theta))d\theta$.

This leads to:

$$= \frac{1}{\sqrt{3}} \int du$$

$$= \frac{1}{\sqrt{3}}u + C$$

Finally, revert back to $x$ by substituting $u = \sec(\theta) - \tan(\theta)$.

Thus, the solution is:

$$= \frac{\sec(\theta) - \tan(\theta)}{\sqrt{3}} + C$$

Since we found $x = \sec(\theta)$, you can rewrite this in terms of $x$:

$$= \frac{x}{\sqrt{3(x^2-1)}} + C$$