Recent content by HappyS5

Is ##(r+y)^3## the same as ##[(1+y/r)^3*r^3]##. If not, how do I factor r out of the parenthesis.

Hello, I have a really basic question. I read the criteria that "see Weinberg, (3.5.13-14)" refers to, and to me, it seems that the equation will always be greater than 0 since it is squared. If so, what is it's purpose? Source...

Remember the integration constant C. Here is what I get:##\frac{1}{3} \int sec^3(\theta) d\theta = \frac{1}{6} (9y \cdot \sqrt{(\frac{1}{3})^2 + y^2}## ##+ \ln \left|3 \big(\sqrt{(\frac{1}{3})^2 + y^2} + y) \right|)##+ C

Thanks. In that case, I would solve it the following way: ##\int \sqrt{(1+(3y)^2)} dy \ and \ 3y = tan(\theta) \ and \ dy = \frac{sec^2(\theta) d\theta}{3}## This gives ##\int \sqrt{(1 + tan^2(\theta))} \cdot \frac{sec^2(\theta)}{3} d\theta## or ## \frac {1}{3} \cdot \int sec^3(\theta)...

I am a bit confused. Why are people substituting ##3y = tan(\theta)## I would substitute ##y=tan(\theta)##, and the deivative is ##\frac{dy}{dx} = tan^2(\theta) + 1 = sec^2(\theta)##. The final setup ## dy = sec^2(\theta) d \theta##. ##\int \sqrt(1 + y^2) dy = \int \sqrt(1 + tan^2(\theta))...