Recent content by Sensayshun

Lol. I'm in integration mode, my bad. Doesn't that just make it even easier? Because that means that: \int{\frac{cost}{1 + sint}dt} = \int{\frac{f'(x)}{f(x)}} therefore: \int{\frac{cost}{1 + sint}} = log(1 + sint) + C?

Just a thought, I'll work through number 4 with substitution as well. But does that not follow the same rule? \int{\frac{cos t}{1 + sin t}dt} Differentiatial of the denominator is -cos t resulting in: -1\int{\frac{cos t}{1 + sin t}dt = -log(1 + sin t) + C? I think I may have gone...

Ahhh ok then. I'll work through these and update tomorrow. Thanks for all of your help again.

So for number 3.) \int{\frac{x.dx}{2 + x^{2}}} I could give the answer 2log(2 + x^{2}) + c? As: \int{\frac{x.dx}{2 + x^{2}}} = 2\int{\frac{f'(x)}{f(x)}}}

Thanks Cyosis. I tried to find a way to give you reputation points but it doesn't seem to exist on this board. Just so you know I really appreciate your help :) For 3.) and 4.) I would've thought substituting the denominator of each fraction. So: u = (2 + x^{2}) and u = (1 + sin t)...

Homework Statement 1.) \int{\frac{t}{(1+t^{2})^3}dt} 2.) \int{xe^{-2x}dx.} 3.) \int{\frac{x.dx}{2+x^{2}}} This one is inegrating between 2 and 0, but I didn't know how to format that in. 4.) \int{\frac{cos t}{1 + sin t}dt.} Homework Equations Q 1.) I'm substituting u =...

To mark, sorry it's meant to be = 0 on the end. Certain bits I couldn't get to format correctly, and I thought it'd be clear(ish). HallsOfIvy - Thanks ever so much, that might make things far simpler! Me and my friends were sat around trying to do this and we thought we'd have gone wrong...

Homework Statement 2iz^2 - (3-8i)z -6 + 7i Homework Equations z = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a} The Attempt at a Solution Right, here goes... a = 2i b = -3 + 8i (is this correct? Or would it be better to leave it as 3-8i?) c = (-6 + 7i) so using these: z =...

Yup sorry, I'd like to blame my stupidity on the hour of the morning, but it's afternoon.

Oh wait, am I being stupid? integral of 1/s wouldn't be the same as 1/f(x) so you'd then use that rule? So then instead of getting zero, which is wrong because surely it isn't zero. That where you use logs? because if you get the answer zero you can't differentiate zero to get back to 1/s...

Ahhh yes, silly me. It would be \frac{-1}{s} not -1/s^2 I have that answer written down so I don't know why I wrote it differently on here :P As for \frac{1}{s} that would integrate to 0s^{0}. And anything to the power zero is 1. But multipled by zero would be zero. So it's nothingness...

That's what I was trying to do, write it as \int s^{-2} which surely then goes to -s^{-1} ? Thanks by the way, very speedy responses.

Homework Statement \int\frac{ds}{s^2} Homework Equations None really...I suppose \int\frac{a}{bx + c} = \frac{1}{b}ln.a.(bx + c) The Attempt at a Solution Pretty much, using the formula above gives you: ln(s$^{2}$) But I was thinking, if you rearrange \frac{ds}{s$^2$}...