1. antiderivate

how do I find the antiderivate to:

cot(x+4)

and

xarccot(4x) ??

2. Originally Posted by weasley74
how do I find the antiderivate to:

cot(x+4)

and

xarctan(4x) ??
$cot(x+4)=\frac{cos(x+4)}{sin(x+4)}$

Set $u=sin(x+4)$
$du=cos(x+4)dx$

$\int \frac{cos(x+4)}{sin(x+4)} dx = \int \frac{du}{u} \Rightarrow ln|u| + C$

Sub $u=sin(x+4)$ back in.

3. I'm sorry! it should be (cot(x+4))^2

4. $\int \cot^2(x+4)~dx$

$\int \frac{\cos^2(x+4)}{\sin^2(x+4)}~dx$

Let $u = x + 4$, then $du = dx$.

$\int \frac{\cos^2 u}{\sin^2 u}~du$

$\int \frac{1 - \sin^2 u}{\sin^2 u}~du$

$\int \frac{1}{\sin^2 u}~du - \int 1~du$

$\int \csc^2 u ~du - \int 1~du$

$-\cot u - u + C$

$-\cot(x+4) - x - 4 + C$

Or simply,

$-\cot(x+4) - x$

$-[x+\cot(x+4)]$

5. Originally Posted by colby2152
$cot(x+4)=\frac{cos(x+4)}{sin(x+4)}$

Set $u=sin(x+4)$
$du=cos(x+4)dx$

$\int \frac{cos(x+4)}{sin(x+4)} dx = \int \frac{du}{u} \Rightarrow ln|u| + C$

Sub $u=sin(x+4)$ back in.
Interesting note, subbing $u=sin(x+4)$ back in gives:

$ln|sin(x+4)| + C$

However, quick integration rules says the integral is:

$-csc^2(x+4) + C$

Setting these equal to each other, since they are equivalent gives:

$ln|sin(x+4)| = -csc^2(x+4)$