# integral of inverse hyperbolic

• Sep 5th 2010, 09:37 PM
integral of inverse hyperbolic
I have to find the integral of (x^2)(arccothx) dx

I have used integration by parts with
u = arccothx
du = 1/(1-x^2)
dv = x^2
v = 1/3*x^3

which i get:

(1/3)(x^3 * arccothx) - integral (x^3)/3(1-x^2)

i am stuck here as I do not know if I have to use polynomial long division on the integrand above or what the next step might be.
• Sep 5th 2010, 11:44 PM
Opalg
Quote:

I have to find the integral of (x^2)(arccothx) dx

I have used integration by parts with
u = arccothx
du = 1/(1-x^2)
dv = x^2
v = 1/3*x^3

which i get:

(1/3)(x^3 * arccothx) - integral (x^3)/3(1-x^2)

i am stuck here as I do not know if I have to use polynomial long division on the integrand above or what the next step might be.

Yes, use polynomial division to reduce the degree of the numerator:

$\displaystyle \dfrac{x^3}{3(x^2-1)} = -\frac13\,\dfrac{x^3}{x^2-1} = -\frac13\Bigl(x +\dfrac x{x^2-1}\Bigr).$
• Sep 5th 2010, 11:48 PM
Prove It
Now try a trigonometric substitution $\displaystyle x = \sin{\theta}$ so that $\displaystyle \theta = \arcsin{\theta}$ and $\displaystyle dx = \cos{\theta}\,d\theta$ and the integral becomes

$\displaystyle \frac{1}{3}\int{\frac{x^3}{1 - x^2}\,dx} = \frac{1}{3}\int{\frac{\sin^3{\theta}}{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}$

$\displaystyle = \frac{1}{3}\int{\frac{\sin^3{\theta}\cos{\theta}}{ \cos^2{\theta}}\,d\theta}$

$\displaystyle = \frac{1}{3}\int{\frac{\sin^3{\theta}}{\cos{\theta} }\,d\theta}$

$\displaystyle = \frac{1}{3}\int{\frac{\sin{\theta}(1 - \cos^2{\theta})}{\cos{\theta}}\,d\theta}$

$\displaystyle = \frac{1}{3}\int{\frac{\sin{\theta} - \sin{\theta}\cos^2{\theta}}{\cos{\theta}}\,d\theta }$

$\displaystyle = \frac{1}{3}\left(\int{\frac{\sin{\theta}}{\cos{\th eta}}\,d\theta} + \int{-\sin{\theta}\cos{\theta}\,d\theta}\right)$

Each of these is solvable using the substitution $\displaystyle u = \cos{\theta}$.
• Sep 6th 2010, 12:23 AM