How do you find the integral of #(2dx) /(x^3sqrt(x^2 - 1))#?

Question

Charles Anctil · Accepted Answer

#int (2dx)/(x^3sqrt(x^2-1)) = arctan(sqrt(x^2-1)) + sqrt(x^2-1)/x^2+C#

For #x in (1,+oo)# substitute:

#x = sect#

#dx = sect tant dt#

with #t in (0,pi/2)#

so:

#int (2dx)/(x^3sqrt(x^2-1)) = 2 int (sect tant dt )/(sec^3tsqrt(sec^2t-1))#

Now:

#sec^2t -1 = tan^2t#

and as for #t in (0,pi/2)# the tangent is positive:

#sqrt(sec^2t -1) = tant#

then:

#int (2dx)/(x^3sqrt(x^2-1)) = 2 int (sect tant dt )/(sec^3t tant)#

#int (2dx)/(x^3sqrt(x^2-1)) = 2 int dt/sec^2t#

#int (2dx)/(x^3sqrt(x^2-1)) = 2 int cos^2tdt#

Now:

#2cos^2t = 1+cos2t#

so:

#int (2dx)/(x^3sqrt(x^2-1)) = int (1+cos2t)dt#

and using linearity:

#int (2dx)/(x^3sqrt(x^2-1)) = int dt +int cos2tdt#

#int (2dx)/(x^3sqrt(x^2-1)) =t + 1/2sin2t+C#

To undo the substitution note that:

#x = sect => tant = sqrt(x^2-1)#

so:

#t = arctan(sqrt(x^2-1))#

and using the parametric fomulas:

#sin 2t = (2tant)/(1+tan^2t) = (2sqrt(x^2-1))/(1+x^2-1) = (2sqrt(x^2-1))/x^2#

So:

#int (2dx)/(x^3sqrt(x^2-1)) = arctan(sqrt(x^2-1)) + sqrt(x^2-1)/x^2+C#

By differentiating we can see that the solution is valid also for #x in (-oo,-1)#

Emilia Aguilar · Answer

undefined To find the integral of $\frac{2dx}{x^3\sqrt{x^2 - 1}}$:

Make the substitution $x = \sec(\theta)$, $dx = \sec(\theta)\tan(\theta)d\theta$.

The integral becomes:
$$
\int \frac{2\sec(\theta)\tan(\theta)d\theta}{\sec^3(\theta)\sqrt{\sec^2(\theta) - 1}}
$$

Simplify the expression:
$$
= \int \frac{2\sec(\theta)\tan(\theta)d\theta}{\sec^3(\theta)\sqrt{\tan^2(\theta)}}
$$
$$
= \int \frac{2\sec(\theta)\tan(\theta)d\theta}{\sec^3(\theta)\tan(\theta)}
$$
$$
= \int \frac{2d\theta}{\sec^2(\theta)}
$$

Now, integrate:
$$
= \int 2\cos^2(\theta)d\theta
$$
$$
= \int (1 + \cos(2\theta))d\theta
$$
$$
= \theta + \frac{1}{2}\sin(2\theta) + C
$$

Substitute back for $\theta$:
$$
= \theta + \sin(\theta)\cos(\theta) + C
$$

Finally, revert back to $x$:
$$
= \cos^{-1}(x) + \sin(\cos^{-1}(x))\cos(\cos^{-1}(x)) + C
$$

Thus, the integral is:
$$
\cos^{-1}(x) + \sqrt{1-x^2}\cdot x + C
$$