Recent content by Jungy

Yeah I have 'I=' in front of my work all the time, just didn't put it here... although you rightly pose an important point.. I'm not 100% confident about freely fiddling with the limits, although I know those 'properties of definite integrals' or whatnot, and are obviously relevant, especially...

dummy variables, solves having to integrate log trigs ^^; And it becomes so simple after all that work, with an easy integration to end. Hahaha I'm so slow... Thanks kof + everyone else who contributed

Ah yes, snipez, silly me, wrote it down and forgot to post it up here it becomes 2/(1+tan (t)); so we get integral ln 2 - ln (1+tan t) but i still don't like the integral ln (1+tan t) bit... how would I go about doing that... with the second approach: If you're saying what I think you're...

Thanks for all the replies + Dick's input (lol) Okay so I'm just going to spell out everything I've tried doing so far: Using kof's substitution -> Expanding tan\,(\frac{\pi}{4}-t) I get \frac{1-tan\,t}{1+tan\,t} The integral of that, I recognised is ln (cos+sin) if that's of any use later...

After a bit of fiddling: \int_0^\frac{\pi}{4} ln[tan(\frac{\pi}{4}-t)+1)\,dt I'm still not picking up anything... by parts from here doesn't look to friendly

Let's go... x = tan\,u dx = sec^2\,u\,du so it becomes \int_0^\frac{\pi}{4} ln[tan(u)+1]\,du and now I'm stuck again...

Homework Statement \int_0^1 \frac{ln(x+1)}{x^2+1} The Attempt at a Solution Alright so I've come across this integral as the last question in the whole exercise set, and... I've gotten nothing... All previous questions have involved some sort of substitution.. but I can't find any...

Ah so what I did do was stupid; made extra work for myself which indeed gave me the wrong answer. Thanks for the explanation.

Yeah I think I got that so if you multiply top and bottom by z^2+1... I get \frac {z(z^2+1)}{(z^2+2)\sqrt{z^2}(z^2+1)^\frac {3}{2}} so everything cancels and I had the root 1+z^2 leftover (incl the z^2 + 2) in the bottom... or am I just doing something plain stupid here? However...

I've done as you've guided me And now I've got (after canceling etc etc) \int \frac {dz}{(z^2+2)(z^2+1)^\frac {1}{2}} What have I done wrong now...? I've got the z^2+2 but I have a leftover (z^2+1)^\frac {1}{2} from the du/dz having the index of 3/2...

Thanks for staying with me here Cyosis; I understand both the substitutions you have provided (by that I mean I know what you're talking about in terms of d/dx etc) but I don't know how I can use the second one. Or is it a straight substitution after the first one you have provided?

Now that I've edited the above... it becomes \frac {1}{a} \theta and \theta = arcsec \frac{x}{a} Yes, that looks right now. Thanks.Edit: However with your second hint, I am still struggling :(

Oops, that's what I meant; not squared... better edit that.

Okay let's see: let x=asec \theta then dx=atan \theta sec \theta Edit: That comes to give \int \frac{1}{a} d\theta That looks right.

1. Question 2 2. \int \frac{dx}{1+cos^2 x} 3. An attempt at a solution I've tried doing various u-substitutions which went nowhere, I've tried making trig identities, but I can't seem to find one that works. I've also tried splitting the 1 dx on top into sin^2 x+cos^2 x etc. but have...