How do you evaluate #int_1^sqrt(3) 4/(x^2sqrt(x^2 - 1)) dx#?

Question

Luke Alvarez · Accepted Answer

The integral has value #4sqrt(2/3)#

Use the substitution #x = sectheta#. Then #dx = secthetatantheta d theta#. You should (stylistically) change the bounds of integration, but for ease of entering on the computer, I haven't. Call the integral #I#.

#I = int_1^sqrt(3) 4/((sec theta)^2sqrt((sec theta)^2 - 1)) secthetatantheta d theta#

# I = int_1^sqrt(3) 4/(sec^2thetasqrt(tan^2theta)) secthetatantheta d theta#

#I = int_1^sqrt(3) 4/(sec^2thetatantheta) secthetatantheta d theta#

#I= int_1^sqrt(3) 4/sectheta d theta#

#I = int_1^sqrt(3) 4costheta d theta#

#I = [4sintheta]_1^sqrt(3)#

DO NOT EVALUATE THIS INTEGRAL. SINCE WE DIDN'T CHANGE THE BOUNDS OF INTEGRATION, THIS INTEGRAL WILL BE INCORRECT. THE CORRECT PROCESS WOULD BE TO SWITCH THE VARIABLE BACK TO #X#.

From our initial substitution, we know that #sectheta = x/1#, so if we were to draw a triangle, the hypotenuse would measure #x# and the side adjacent #theta# would measure #1#. By pythagoras, the side opposite would measure #sqrt(x^2 - 1)#. Therefore, #sintheta = sqrt(x^2- 1)/x#.

#I = [(4sqrt(x^2 - 1))/x]_1^sqrt(3)#

#I = (4sqrt(sqrt(3)^2 - 1))/sqrt(3) - (4sqrt(1^2 - 1))/1#

#I = (4sqrt(2))/sqrt(3) - 0#

#I = 4sqrt(2/3)#

Hopefully this helps!

Scarlett Allison · Answer

undefined To evaluate the integral $ \int_1^{\sqrt{3}} \frac{4}{x^2\sqrt{x^2 - 1}} \, dx $, we first notice that the integrand resembles the derivative of the inverse trigonometric function. Specifically, the integrand is similar to the derivative of $ \sec^{-1}(x) $, which is $ \frac{1}{x\sqrt{x^2 - 1}} $.

We can manipulate the integrand to make it resemble the derivative of $ \sec^{-1}(x) $ by multiplying and dividing by $ \sqrt{x^2 - 1} $:

$$
\begin{align*}
\int_1^{\sqrt{3}} \frac{4}{x^2\sqrt{x^2 - 1}} \, dx &= \int_1^{\sqrt{3}} \frac{4 \cdot \sqrt{x^2 - 1}}{(x^2\sqrt{x^2 - 1}) \cdot \sqrt{x^2 - 1}} \, dx \
&= \int_1^{\sqrt{3}} \frac{4\sqrt{x^2 - 1}}{(x\sqrt{x^2 - 1})^2} \, dx
\end{align*}
$$

Now, let $ u = x\sqrt{x^2 - 1} $, then $ du = \frac{x^2}{\sqrt{x^2 - 1}} \, dx $. We can rewrite the integral in terms of $ u $:

$$
\int_1^{\sqrt{3}} \frac{4\sqrt{x^2 - 1}}{(x\sqrt{x^2 - 1})^2} \, dx = \int_{\sqrt{2}}^{\sqrt{6}} \frac{4}{u^2} \, du
$$

This integral can be easily evaluated:

$$
\int_{\sqrt{2}}^{\sqrt{6}} \frac{4}{u^2} \, du = \left[ -\frac{4}{u} \right]_{\sqrt{2}}^{\sqrt{6}} = -\left( \frac{4}{\sqrt{6}} - \frac{4}{\sqrt{2}} \right) = -\frac{4}{\sqrt{6}} + 2\sqrt{2}
$$

Therefore, $ \int_1^{\sqrt{3}} \frac{4}{x^2\sqrt{x^2 - 1}} \, dx = -\frac{4}{\sqrt{6}} + 2\sqrt{2} $.