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

2 Answers
May 12, 2018

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

Explanation:

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))^') #

Next is IBP:

#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 #

May 12, 2018

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

Explanation:

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 #