Bonus Homework Problem: Integral of csc

[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.

 

[lanedance]

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

 

[Mark44]

\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)

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.

\sqrt{}{u²-1} = 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.

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

 

[HallsofIvy]

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).