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try it out and if you don't like it u can quickly drop

i don't

he's talking abt someone else

wassup @NoctNight

ye he's ass bro

but ye my teacher said such questions aren't likely to come up in hsc cos could be controversial

eh i have seen such questions in trials (killara iirc) and have been called out for using it by other ppl LOL

expanding on this, would just say that such a method just isn't rigourous but if there are applicable qs in the hsc u would use it cos they don't rlly care abt this stuff.

not too sure abt this either, but i think it probably just isn't properly defined yet?

I was arguing that the answers u obtain using these methods aren't the same function because of the inherent restriction u put on certain values, which can cause problems such as what u pointed out in evaluating definite integrals. I am also fully aware of the trick, in fact I've used it before...

What I mean is just because the original integrand is undefined doesn't mean the primitive is also undefined at that point. This is because the primitive could have a vertical tangent at the point where the original function is undefined (e.g. 1/sqrt(1-x^2) is undefined at x = +-1 yet it's...

this is irrelevant since the original function being undefined at a point doesn't mean it's primitive is also undefined, eg f(x) = 1/(sqrt(1-x^2)). In this case the manipulation works (I think) and the results end up being the same due to how we define trig functions most likely, but as a...

i agree i did make a silly mistake

nice, i was intending for it to be done by sqrt(tanx) = 1/2 (sqrttanx + sqrtcotx + sqrttanx - sqrtcotx) tho

but when u divide both sides by u^2, a problem is gonna be created when u = 0 so the answers are not actually the same

Now for a follow up question: hence or otherwise integrate sqrt(tanx):)

haven't asked them yet (except in the sample paper) so they could realistically ask them but it'll be at most 1 question in the paper

thanks:)

Well if u do it without the limits, then after u do the sub u cannot just use a dummy variable to change the integral from in terms of u back to in terms of x. Thus when u integrate by parts the first integral u can't just cancel out the integrals cos one of them is in terms of u and the other...

the differentiation calculator says it's not, I think this is one of those integrals where u have to rely on the bounds being a certain way to properly solve it.