Question

How do you integrate `int xsec^-1x` by integration by parts method?

Answer 1

This is a problem that will necessitate integration by parts. We let `u = sec^-1x` and `dv = x`. We can easily see that `v = 1/2x^2`, but we will have to work to find the derivative of `u`.

`u = sec^-1x`

`secu = x`

`secutanu(du)/dx= 1`

`(du)/dx= 1/(secutanu)`

Because `x = secu`, this means that the side adjacent `u`, if we were to draw an imaginary triangle, would measure `1` and the hypotenuse would measure `x`.

This means the side opposite would measure `sqrt( x^2-1)`, and so `tanu = sqrt(x^2-1)/ 1= sqrt(x^2 - 1)`.

`(du)/dx = 1/(xsqrt(x^2 - 1)`

`du = 1/(xsqrt(x^2 - 1)) dx`

We now have, by the integration by parts formula,

`int(u dv) = uv - int(v du)`

`intxsec^-1xdx = sec^-1x(1/2x^2) - int1/2x^2 1/(xsqrt(x^2 - 1))dx`

`intxsec^-1xdx = sec^-1x1/2x^2 - 1/2 int x/sqrt(x^2 - 1)dx`

We could use trig substitution to integration `x/sqrt(x^2 - 1)`. However, in this case, a u-substitution would do the trick easily.

Let `u = x^2 - 1`. Then `du = 2xdx -> dx = (du)/(2x)`.

`intxsec^-1xdx = sec^-1x1/2x^2 - 1/2int x/sqrt(u) * (du)/(2x)`

`intxsec^-1xdx = sec^-1x1/2x^2 - 1/4 int u^(-1/2) du`

`intxsec^-1x dx = 1/2x^2sec^-1x - 1/4(2u^(1/2)) + C`

`intxsec^-1xdx = 1/2x^2sec^-1x - 1/2u^(1/2) + C`

`intxsec^-1xdx = 1/2x^2sec^-1x - 1/2(x^2 - 1)^(1/2) + C`

Hopefully this helps!