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

1 Answer
Mar 3, 2017

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!