A Tip

[BigGlenntheHeavy]

$\displaystyle This \ is \ in \ regards \ to \ Aladdin's \ query \ below.$

$\displaystyle Hint: \ A \ tip, \ b \ a \ constant, \ obviously \ b \ \ne \ 0.$

$\displaystyle \int\frac{du}{u^{2}-b} \ \implies \ Natural \ Logarithm \ Solution, \ \frac{1}{u^{2}-b} \ can \ be \ expanded.$

$\displaystyle \int\frac{du}{u^{2}+b} \ \implies \ Arctangent \ Solution, \ \frac{1}{u^{2}+b} \ is \ prime.$

 

[Aladdin]

Thanks for that , Glenn .

 

[Deleted member 4993]

Anotherway

$\displaystyle \int\frac{du}{u^2+b^2} = \frac{i}{2b}ln\left |\frac{u+ib}{u-ib}\right |+ C$

This representation is useful in complex analysis.

 

[galactus]

Yes, indeed. There are literally millions of these identities in some form or another.

Here's one, we do not see very often, they I like. The clever use of which may prove useful in solving various limits or what not.

$\displaystyle \frac{sin(x)}{x}=cos(\frac{x}{2})\cdot cos(\frac{x}{4})\cdot cos(\frac{x}{8})\cdot\cdot\cdot$

$\displaystyle \frac{sin(x)}{x}=\prod_{k=1}^{\infty}cos(\frac{x}{2^{k}})$

which means that $\displaystyle \lim_{n\to \infty}2^{n}sin(\frac{x}{2^{n}})=x$

Here is one for arctan related to SK's, it would appear.

$\displaystyle tan^{-1}(x)=\frac{i}{2}\cdot ln(\frac{i+x}{i-x})$.