# Bonus Homework Problem: Integral of csc

1. Feb 25, 2009

### ƒ(x)

$$\int csc dx = ?$$

I just am seeking confirmation as to whether or not I did this correctly. I apologize beforehand for any formatting errors.

$$\int csc(x) dx =$$ $$\int 1/sin(x) dx$$
u = sin(x)
du = cos(x)dx
du / cos(x) = dx

u2 = sin2(x)
u2 = 1 - cos2(x)
$$\sqrt{}$${u^2-1} = cos(x)
du / $$\sqrt {u2-1} = dx$$

Therefore

$$\int du / u\sqrt{}$${u2-1}

And, since I cannot remember the arc property, this is as for as I was able to get.

Last edited by a moderator: Feb 26, 2009
2. Feb 25, 2009

### lanedance

try u = ln(tan(x/2))

3. Feb 26, 2009

### Staff: Mentor

This is an interesting approach, but you have a problem in the next line. The intermediate step is: u2 - 1 = - cos2(x).
The right side is always <= 0, so taking its square root almost always gives you imaginary numbers.

That's an interesting approach, but there's a problem here:

4. Feb 26, 2009

### HallsofIvy

Staff Emeritus
I would integrate $\int dx/cos(x)$ by multiplying both numerator and denominator by cos(x): $\int cos(x)dx/cos^2(x)= \int cos(x)dx/(1- sin^2(x))$ and then let u= sin(x).