Oliver Spryn
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 Nov5 asked Combinatorics: When To Use Different Counting Techniques Oct17 awarded Popular Question Oct11 comment Pythagorean theorem: $A^2 + A^2 = C^2$ How to solve for $A$? @Campbeln Not a problem! If you found my answer helpful, could you accept the answer, please? Oct11 accepted How is $2 + e^{t^2} - e^{-t^2}$ Equal to $(e^t + e^{-t})^2$? Oct11 comment How is $2 + e^{t^2} - e^{-t^2}$ Equal to $(e^t + e^{-t})^2$? Oh... Thank you for making that clear. The \frac{e^t}{e^t} cancel to give one. That is clear in the way you broke it up. Unfortunately, I was given the longer expression to start with, and had to see if it simplified, so I didn't have the opportunity to work backwards. Oct11 comment How is $2 + e^{t^2} - e^{-t^2}$ Equal to $(e^t + e^{-t})^2$? @AustinMohr Yep, I made a silly typo. Oct11 revised How is $2 + e^{t^2} - e^{-t^2}$ Equal to $(e^t + e^{-t})^2$? added 15 characters in body Oct11 answered Pythagorean theorem: $A^2 + A^2 = C^2$ How to solve for $A$? Oct11 asked How is $2 + e^{t^2} - e^{-t^2}$ Equal to $(e^t + e^{-t})^2$? Aug15 awarded Autobiographer Mar15 comment Divide with remainder $\frac{x^2}{x^2 + x + 2}$ Woo-hoo! Just what I was looking for! Thank you, Arturo! Mar15 accepted Divide with remainder $\frac{x^2}{x^2 + x + 2}$ Mar14 accepted Why Doesn't This Integral $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$ Work? Mar14 accepted Equivalent to \begin{align}\int \cos{\left(2x\right)} \ dx\end{align}? Mar14 asked Divide with remainder $\frac{x^2}{x^2 + x + 2}$ Mar14 accepted Find $\int e^{2\theta} \cdot \sin{3\theta} \ d\theta$ Mar14 comment Equivalent to \begin{align}\int \cos{\left(2x\right)} \ dx\end{align}? Thank you all!! Mar14 comment Equivalent to \begin{align}\int \cos{\left(2x\right)} \ dx\end{align}? @anon You'd have to dig around to find what I'm references. Mar14 asked Equivalent to \begin{align}\int \cos{\left(2x\right)} \ dx\end{align}? Mar12 comment Integration: How to Begin? Hmm... how can we stop thinking in terms of right triangles if the above example uses $\tan{x}$ and $\cos{x}$? Plus, by having a square root, we easily know that it would (in this case) exist as the hypotenuse of the triangle.