Recent content by Electrifying

No they were written as arcsin and arccos. It was just the hyperbolic ones that were written without a c.

Thanks, I did the second step of the parts with the wrong inspection method (used the log method instead of the raise the power and check co-efficient method). I'm now at the right answer, thanks for your help everyone, and thanks for that website too. Also about the arcsinh thing, we've been...

Thanks for that, I got the following final answer: \frac{1}{2}. (x^2 .arsinh(x^2) - ln(1+x^4)^{0.5}) +C Oh and why let u=arcsin(x) for the sub? I assume you meant arsinh and just misread?

Hi guys, I don't even know where to begin with this question. Find the following indefinite integral: \int x.arsinh (x^2) dx Thanks very much for any help, its much appreciated.

Thank you very much for that, my final answer of \frac{dy}{dx} =-\frac{x}{\sqrt{a^2-x^2}} Agrees with the answer obtained if you just differentiate 'normally'. Thanks again!

Sorry, just a typo, b^2 should be x^2. But a is just a constant yes.

Hey guys, I've got the following 2 mark question on a problem sheet, but I can't seem able to do it. I'd appreciate any help, thanks. Differentiate, from first principles, the following: y=\sqrt(a^2-x^2) I know I have to take the limit as δx tends to 0 of [(f(x+δx)- f(x)]/δx but can't seem to...