Difference between revisions of "Antiderivative of arctan"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
| − | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log \left( | + | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log \left(z^2+1 \right)+C,$$ |
where $\mathrm{arctan}$ denotes the [[arctan|inverse tangent]] and $\log$ denotes the [[logarithm]]. | where $\mathrm{arctan}$ denotes the [[arctan|inverse tangent]] and $\log$ denotes the [[logarithm]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | Let $u=\mathrm{arctan}(z)$ and $\mathrm{d}v=1$. Then $v=z$ and because of the [[derivative of arctan]], $\mathrm{d}u=\dfrac{1}{z^2+1} \mathrm{d}z$. Therefore using [[integration by parts]], | |
| − | + | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z \mathrm{arctan}(z) - \displaystyle\int \dfrac{z}{z^2+1} \mathrm{d}z.$$ | |
| + | To evaluate the remaining integral, let $w=z^2+1$ so that $\mathrm{d}w=2z \mathrm{d}z$, i.e. $\dfrac{1}{2} \mathrm{d} w = z \mathrm{d}z$ and we see using the [[derivative of the logarithm]], | ||
| + | $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z \mathrm{arctan}(z) - \dfrac{1}{2}\displaystyle\int \dfrac{1}{w} \mathrm{d}w = z \mathrm{arctan}(z) - \dfrac{1}{2} \log(z^2+1) + C,$$ | ||
| + | as was to be shown. | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 12:52, 10 October 2016
Theorem
The following formula holds: $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z\mathrm{arctan}(z) - \dfrac{1}{2}\log \left(z^2+1 \right)+C,$$ where $\mathrm{arctan}$ denotes the inverse tangent and $\log$ denotes the logarithm.
Proof
Let $u=\mathrm{arctan}(z)$ and $\mathrm{d}v=1$. Then $v=z$ and because of the derivative of arctan, $\mathrm{d}u=\dfrac{1}{z^2+1} \mathrm{d}z$. Therefore using integration by parts, $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z \mathrm{arctan}(z) - \displaystyle\int \dfrac{z}{z^2+1} \mathrm{d}z.$$ To evaluate the remaining integral, let $w=z^2+1$ so that $\mathrm{d}w=2z \mathrm{d}z$, i.e. $\dfrac{1}{2} \mathrm{d} w = z \mathrm{d}z$ and we see using the derivative of the logarithm, $$\displaystyle\int \mathrm{arctan}(z) \mathrm{d}z = z \mathrm{arctan}(z) - \dfrac{1}{2}\displaystyle\int \dfrac{1}{w} \mathrm{d}w = z \mathrm{arctan}(z) - \dfrac{1}{2} \log(z^2+1) + C,$$ as was to be shown.
