# Thread: calculus proof

1. ## calculus proof

Hi, Ive been stuck on this proof problem for awile and was wondering if anyone has any ideas?

Integral (csc(x)) dx = ? expain how you get the answer.

**For example in Integral (sec(x)) dx:

Integral (Sec(x)) (tan(x) + sec(x)) / (tan(x) + sec(x))) dx =

(sec(x)tan(x) + sec^2(x)) / (tan(x) + sec (x))

Set u= tan(x) + sec(x) du= (sec^2(x) + sec(x)tan(x)) dx

If, Integral sec(x) = Integral (1 / u) = ln(abs(u))

Then, Integral sec(x) = ln(abs(tan(x) + sec(x))) + c

With the csc(x) proof I am stuck trying to find the function equaling one to times it by in the beginning that will lead to the correct answer.

2. $\displaystyle \int \csc{x} \, dx$

multiply by $\displaystyle \frac{\csc{x} + \cot{x}}{\csc{x} + \cot{x}}$

$\displaystyle \int \frac{\csc{x}(\csc{x} + \cot{x})}{\csc{x} + \cot{x}} \, dx$

$\displaystyle \int \frac{\csc^2{x} + \csc{x}\cot{x}}{\csc{x} + \cot{x}} \, dx$

$\displaystyle -\int \frac{-\csc^2{x} - \csc{x}\cot{x})}{\csc{x} + \cot{x}} \, dx$

$\displaystyle -\ln|\csc{x} + \cot{x}| + C$

$\displaystyle \ln \left| \frac{\sin{x}}{1 + \cos{x}} \right| + C$

3. Hello, ulion!

The routine is similar to that of: $\displaystyle \int\sec x\,dx$

$\displaystyle \int \csc x\,dx$

Multiply by $\displaystyle \frac{\csc x - \cot x}{\csc x - \cot x}$

$\displaystyle \int\left( \csc x\cdot\frac{\csc x - \cot x}{\csc x - \cot x}\right)\,dx \;\;=\;\;\int \frac{\csc^2\!x - \csc x\cot x}{\csc x-\cot x}\,dx$

. . $\displaystyle = \;\;\int\frac{\overbrace{-\csc x\cot x + \csc^2x}^{du}}{\underbrace{\csc x - \cot x}_u}\,dx \;\;=\;\;\ln|\csc x - \cot x| + C$

4. thankyou for your help