Difference between revisions of "Arctan"

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<strong>Proposition:</strong>  
 
<strong>Proposition:</strong>  
$\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C$
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$\displaystyle\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C$
 
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<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  

Revision as of 05:58, 31 October 2014

The $\mathrm{arctan}$ function is the inverse function of the tangent function.

Properties

Proposition: $\dfrac{d}{dz} \mathrm{arctan}(z) = \dfrac{1}{z^2+1}$

Proof: If $\theta=\mathrm{arctan}(z)$ then $\tan \theta = z$. Now use implicit differentiation with respect to $z$ yields $$\sec^2(\theta)\theta'=1.$$ The following triangle shows that $\sec^2(\mathrm{arctan}(z))=z^2+1$:

Sec(arctan(z)).png

Substituting back in $\theta=\mathrm{arccos(z)}$ yields the formula $$\dfrac{d}{dz} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1}. █$$

Proposition: $\displaystyle\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C$

Proof:

Proposition: $\mathrm{arctan}(z) = \mathrm{arccot}\left( \dfrac{1}{z} \right)$

Proof:

References

Weisstein, Eric W. "Inverse Tangent." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/InverseTangent.html