Question

What is the integral of `int (csc2x)dx`?

Answer 1

`int csc(2x) "d"x = -1/2 ln|csc(2x) + cot(2x)| + "constant"`

`int csc(2x) "d"x = 1/2 ln|tan(x)| + "constant"`

Both answers are equivalent. Differentiate them if you have any doubt.

Explanation:

First, use the `sin` double angle formula `sin2theta -= 2sinthetacostheta` and the Pythagorean Identity `sin^2theta + cos^2theta -= 1`.

`int csc(2x) "d"x = int 1/sin(2x) "d"x`

`= int 1/(2sin(x)cos(x)) "d"x`

`= int cos(x)/(2sin(x)cos^2(x)) "d"x`

`= 1/2 int cos(x)/(sin(x)(1-sin^2(x))) "d"x`

Next, substitute `u = sin(x)`.

`frac{"d"u}{"d"x} = cos(x)`

`frac{"d"u}{"d"x} * "d"x = cos(x)*"d"x`

`1/2 int cos(x)/(sin(x)(1-sin^2(x))) "d"x = 1/2 int 1/(u(1-u^2)) "d"u`

Next write the partial fractions.

`1/2 int 1/(u(1-u^2)) "d"u = 1/2 int 1/u "d"u`

`- 1/4 int 1/(1+u) "d"u - 1/4 int 1/(u-1) "d"u`

`= 1/4 ln|u^2/(1-u^2)| + "constant"`

Now revert `u` back to `sin(x)`.

`1/4 ln|u^2/(1-u^2)| = 1/4 ln|sin^2(x)/(1-sin^2(x))|`

`= 1/4 ln|sin^2(x)/cos^2(x)|`

`= 1/4 ln|tan^2(x)|`

`= 1/2 ln|tan(x)|`