Question

How do you find `\int ( 4- 3\cos x ) ^ { \frac { 1} { 2} } d x`?

Answer 1

Use the trigonometric identity:

`1-cosx = 2sin^2(x/2)`

to have:

`int sqrt(4-3cosx)dx = int sqrt (1+3(1-cosx))dx = int sqrt(1+6sin^2(x/2))dx`

Substitute now:

`t = sin(x/2)`

`x = 2arcsint`

`dx = (2dt)/sqrt(1-t^2)`

so:

`int sqrt(4-3cosx)dx = 2 int sqrt(1+6t^2)/sqrt(1-t^2) dt`

This is an elliptic integral of the second kind and as thus cannot be expressed in terms of elementary functions.