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