Question

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

Answer 1

`= 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`

Answer 2

`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`