How do you integrate #int x arcsec x # using integration by parts?

Question

Charles Anctil · Accepted Answer

# = x^2/2 \ arcsec x - 1/2 sqrt(x^2 - 1) + C #

Clearly, you must need to know #(arcsec(x))^'# to do this by IBP.

So let #y = arcsec(x)#:

#sec y = x#

#(sec y = x)^' implies sec y \ tan y \ y' = 1#

#implies y' = 1/(sec y \ tan y) = 1/(x sqrt(x^2-1)) color(red)(= (arcsec(x))^') #

IBP comes next:

#int f'(x) g(x) \ dx = f(x) \ g(x) - int f(x) g'(x) \ dx#

#implies int dx \ x \ arcsec x = int dx \ (x^2/2)^' \ arcsec x#

#= x^2/2 \ arcsec x - int dx \ x^2/2 \ 1/(x sqrt(x^2-1))#

#= x^2/2 \ arcsec x - 1/2 int dx \ x/( sqrt(x^2-1)) = triangle#

Given that: #(sqrt(x^2 - 1))^' = 1/2 * 1/( sqrt(x^2-1))* 2x = x/( sqrt(x^2-1))#

Then:

#triangle = x^2/2 \ arcsec x - 1/2 int dx \ (sqrt(x^2 - 1))^' #

# = x^2/2 \ arcsec x - 1/2 sqrt(x^2 - 1) + C #

Evelyn Agard · Answer

# int \ x \ "arcsec" \ x \ dx = 1/2x^2 \ "arcsec" \ x - 1/2 sqrt(x^2-1) + c #

We seek: # I = int \ x \ "arcsec" \ x \ dx # We can then apply Integration By Parts: Let # { (u,="arcsec" \ x, => (du)/dx,=1/(xsqrt(x^2-1))), ((dv)/dx,=x, => v,=1/2x^2 ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx # We have: # int \ ("arcsec" \ x)(x) \ dx = ("arcsec" \ x)(1/2x^2) - int \ (1/2x^2)(1/(xsqrt(x^2-1))) \ dx # # I = 1/2x^2 \ "arcsec" \ x - 1/2 \ int \ x/(sqrt(x^2-1)) \ dx # For the integral IBP has introduced, we can perform a substitution, Let: # u = x^2-1 => (du)/dx = 2x # And if we substitute this into the integral we get: # int \ x/(sqrt(x^2-1)) = int \ (1/2)/sqrt(u) \ du # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1/2 \ int \ u^(-1/2) \ du # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1/2 \ (u^(1/2))/(1/2) # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = sqrt(u) # And restoration of the substitution gives us: # int \ x/(sqrt(x^2-1)) = sqrt(x^2-1) # And combining our results, we get: # I = 1/2x^2 \ "arcsec" \ x - 1/2 sqrt(x^2-1) + c #

Cameron Alexander · Answer

undefined To integrate $ \int x \sec^{-1}(x) \, dx $ using integration by parts, we let $ u = \sec^{-1}(x) $ and $ dv = x \, dx $. Then, $ du = \frac{1}{|x|\sqrt{x^2 - 1}} \, dx $ and $ v = \frac{1}{2}x^2 $. Applying the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

We get:

$$ \int x \sec^{-1}(x) \, dx = \frac{1}{2}x^2 \sec^{-1}(x) - \frac{1}{2} \int \frac{x^2}{|x|\sqrt{x^2 - 1}} \, dx $$

To evaluate the remaining integral, we can let $ t = \sqrt{x^2 - 1} $, so $ x = \cosh(t) $ and $ dx = \sinh(t) \, dt $. Substituting:

$$ \int \frac{x^2}{|x|\sqrt{x^2 - 1}} \, dx = \int \frac{\cosh^2(t)}{\cosh(t)} \sinh(t) \, dt $$

$$ = \int \cosh(t) \sinh(t) \, dt $$

$$ = \frac{1}{2} \sinh^2(t) + C $$

Finally, substituting back $ t = \sec^{-1}(x) $, we get:

$$ \int x \sec^{-1}(x) \, dx = \frac{1}{2}x^2 \sec^{-1}(x) - \frac{1}{2} \left( \frac{1}{2} \sinh^2(\sec^{-1}(x)) \right) + C $$