Trig Integral

**dbakeg00** · December 1st 2010, 05:55 PM

Here is my original problem:

$\displaystyle\int 2\tan^{-1}(\frac{z}{4})dz$

$\displaystyle u=2tan^{-1}(\frac{z}{4})$

$\displaystyle du=\frac{2*\frac{1}{4}}{1+(\frac{z}{4})^2}dz=\frac {\frac{1}{2}}{1+\frac{1}{16}z^2}dz$

$\displaystyle dv=dz$ and $\displaystyle v=z$

$\displaystyle\int 2\tan^{-1}(\frac{z}{4})dz=2\tan^{-1}(\frac{z}{4})-\int \frac{\frac{1}{2}z}{1+\frac{1}{16}z^2}dz=2\tan^{-1}(\frac{z}{4})-ln(1+\frac{1}{16}z^2)+c$

Here is what the book got:

$\displaystyle =2\tan^{-1}(\frac{z}{4})-ln(1+\frac{1}{4}z^2)+c$

I know I must have made algebra mistake somewhere, I just cant figure it out where I went wrong. Please help! Thanks.

**dwsmith** · December 1st 2010, 06:05 PM

First, you forgot to multiply my z in the first section. uv-int vdu.

**dwsmith** · December 1st 2010, 06:09 PM

Also, you should have made a second u sub in the vdu integral.

$\displaystyle 2 z*tan^{-1}\left(\frac{z}{4}\right)-2\int\frac{4}{z^2+16}dz$

$\displaystyle u=z^2+16 \ du=2zdz\rightarrow \frac{1}{2}du=zdz$

$\displaystyle 2 z*tan^{-1}\left(\frac{z}{4}\right)-\frac{8}{2}\int\frac{du}{u}\rightarrow ....$

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