Expert: Socrates - 12/1/2006

Question
How do you solve problems such as the integral from 2 to 4 of x(arc secx)dx? I never understood any of the arc sec or arccos type of problems.

Answer
The important thing to know is that the inverse trig functions have algebraic derivatives, in other words, the derivatives are just expressions using polynomials, fractions and radicals. As a consequence of this, the integrals can often be done by integration by parts. But, of course, you must know the derivatives of the inverse trig functions. I will assume you know these.

The derivative of arccosx = -1/(1-x^2)^1/2

The derivative of arcsecx = 1/(|x|(x^2-1)^1/2) =
1/((x)(x^2-1)^1/2) , because x is positive in your example



For example , to find S arccosx dx

Write arccosx = (1)(arccosx)

Use integration by parts and get

S arccosx dx  = (x)(arccosx) - S (x) (-1)/(1-x^2)^1/2 dx =

(x)(arccosx) -   S -x/(1-x^2)^1/2 dx  

This integral is easy to do by substitution, let u = 1-x^2

so du/dx = -2x  and dx = - du/2x

The integral becomes S -x/u^1/2 (-du/2x) =
S (1/2)(u^-1/2) du = u^1/2

So we get S arccosx dx = (x)(arccosx) - (1-x^2)^1/2


For S arcsecx dx , use integration by parts again in the same way

Write arcsecx = (1)(arcsecx)

so S arcsecx dx = (x)(arcsecx) - S x/((x)(x^2-1)^1/2)dx =

(x)(arcsecx) - S 1/(x^2-1)^1/2 dx

To find S 1/(x^2-1)^1/2 dx , let x = secv , so
dx/dv = tanv secv  and dx = tanv secv dv

The integral becomes

S 1/((secv)^2 -1)^1/2 tanv secv dv =

S 1/((tanv)^2)^1/2 tanv secv dv

S 1/tanv tanv secv dv =

S secv dv =

ln(secv + tanv)

now substitute back v = arcsecx

= ln (x + (x^2-1)^1/2)

so the answer is


S arcsecx dx = (x)(arcsecx) - ln (x + (x^2-1)^1/2)