Integrating by Parts: tan-1x dx

[Col Musstard]

Homework Statement

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

Homework Equations

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?

 

[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}

 

[Col Musstard]

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

 

[Col Musstard]

i can't seem to get this problem to work, it seems to repeat itself

 

[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.