# Integrate by parts

1. Feb 25, 2010

### Col Musstard

1. The problem statement, all variables and given/known data

integral tan-1x dx
i am supposed to integrate this by parts

2. Relevant equations

3. The attempt at a solution
integral tan-1x dx = integral cosx/sinx dx
u=cos x, du=-sin x dx
v=ln sin x, dv= sin-1x dx
integral cosx/sinx dx= cosx ln(sinx) - integral[ ln(sinx)(-sinx) dx]
is this correct so far?

Last edited: Feb 25, 2010
2. Feb 25, 2010

### shallgren

Careful with your terminology, the problem is vastly different if its tan^-1(x) (arctangent) or (tan(x))^-1 (cotangent).

If it's the inverse function, the correct fraction would be:
$$\arctan{x}= \frac{{\arcsin{x}}}{\arccos{x}}$$

If it's cotangent, I would set the problem up so that you have $$\int{\cos{x} \cdot \csc{x}\,dx}$$ so that you have the form $$\int{U \cdot V}$$

Last edited: Feb 26, 2010
3. Feb 25, 2010

### Col Musstard

well, it is arctan x so i need to do some recalculating now

Last edited: Feb 25, 2010
4. Feb 25, 2010

### Col Musstard

i cant seem to get this problem to work, it seems to repeat itself

5. Feb 26, 2010

### shallgren

Ah, sorry I just worked it out and it seems that breaking arctan up wasn't the way to go. Instead, do the following:

$$\int{\arctan{x} \cdot 1 \,dx}$$

This way, you can differentiate arctan and integrate 1 without having to repeat integration by parts as you would with breaking up the arctan. Let me know if you need more help.