Submitted by TR14 on Fri, 02/24/2017 - 18:59
Slight issue with Q2 (ii) : I understand how the question's supposed to be answered but I'd like to know why my method doesn't work :
We have $\int_0^{2\pi} \frac{x\sin x}{3 + \sin^2x} dx$, substituting x=2u gives us $$4\int_0^\pi \frac{u\sin(2u)}{3+\sin^2(2u)}du = 4\int_0^\pi \frac{2u\cos u\sin u}{3+(2\cos u\sin u)^2}du$$ My idea was to try and use the simplification given in the question on this integral, where we could write $\int_0^\pi xf(\sin x)dx = \frac12\pi\int_0^\pi f(\sin x)dx$, here using $$f(x)=\frac{2x\cos x}{3+(2x\cos x)^2}$$ giving us $$4\int_0^\pi \frac{u\sin(2u)}{3+\sin(2u)^2}du = 2\pi\int_0^\pi \frac{2\sin(2u)}{3+\sin^2(2u)}du$$ This basically removes the u factor in the numerator, and enables us to proceed just as in (i). However doing so leads us to find that the integral is equal to zero, does this mean that we can't apply the simplification in this case, or is there a mistake somewhere ?
This question is often a
This question is often a source of difficulty for many students. In your case, the prime confusion seems to come from your overloading of $x$. Your $f(x)$ doesn't quite work here. Note that if you let $$f(y) = \frac{2y\cos y}{3 + (2y\cos y)^2}$$ then $$f(\sin x) = \frac{2\sin x\cos(\sin x)}{3 + (2y\cos(\sin x))^2}$$ which really isn't what you had before.
In loose terms, $f(\sin x)$ means that whenever a variable occurs in the expression of the function, it occurs in the form of $\sin x$. So $f(y) = y + a$ can't really be put in the form $f(\sin x)$.
In your case, you could potentially rectify your problem by writing $\cos x = \pm\sqrt{1-\sin^2 x}$ but that would give you quite a painful integral. Can you see why you're going wrong now? Let me know if you don't, I'd be happy to elucidate further.
Didn't see your reply in time
Indeed, seems that one way to overcome this is by considering $\cos(x)=\pm\sqrt{1-sin(x)}$ but then again you have to split up the integral...
Indeed: the straightforward
Indeed: the straightforward way to tackle this bit of the question is to split the integral into $\int_{[0, \pi]} + \int [\pi, 2\pi]$ and then use $x \mapsto \pi - x$ on the latter.
Mistake
The stem result makes use of the fact that $\sin(x)=sin(\pi-x)$, and what actually happens when trying to apply it to $$\int_{0}^{\pi} \frac{u \sin(2u)}{3+sin^2(2u)} du$$ is that making the substitution $u= \pi - t$, gives $\sin(2u)=\sin(2\pi-2t)=-\sin(2t)$ and the integral becomes $$-\int_{0}^{\pi} \frac{(\pi-t)sin(2t)}{3+sin^2(2t)} dt$$ equalizing results in $$\int_{0}^{\pi} \frac{u \sin(2u)}{3+sin^2(2u)} du = \int_{0}^{\pi} \frac{(u-\pi)sin(2u)}{3+sin^2(2u)} du $$ the integral we wanted cancels, remaining just with $$\int_{0}^{\pi} \frac{sin(2u)}{3+sin^2(2u)} dt = 0$$
?
I think the OP knows that $\int_{[0, \pi]} \sin 2u / (3 + \sin^2 2u) \, \mathrm{d}u =0 $ but I can't see how this answer the question at all. The problem with your method is exactly what you said "the integral we wanted cancels".
I just wanted
I just wanted to point out what the mistake was.
Yes, I may be being dense,
Yes, I may be being dense, but I think you've only shown the correct way to evaluate the integral in question and not actually what the mistake was.
I'm sorry
I'm sorry, I think my phrasing was bad. What I wanted to say was that by trying to use the stem result and not paying attention to the fact that $\sin(2\pi-2x)=-\sin(2x)$, he arrived at something that didn't really help.
Ah I see, cheers.
Ah I see, cheers.