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I came across this problem which says:

Let $f:[-1,1]\rightarrow \ \mathbb{R}$ be continuous. Assume that $\int_{-1}^{1}f(t)\, dt =2$. Then $$\lim_{n\to\infty} \int_{-1}^{1}f(t)\sin^2(nt)\,dt$$ equals to

a) $0$

b) $1$

c) $f(1) - f(-1)$

d) Does not exist

I have taken $f(t)=1$ so that it satisfies the given definite integral. Then I see the solution to be $1$. Am I correct? I am looking for a better way to solve it. Please help.

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    I don't get it. We know that $\int_{-1}^{1}f(t)dt =2$ and we want to find the limit of $\int_{-1}^{1}f(t)dt$. The limit of a constant? The thing under the $\lim$ sign should be a function or a sequence, otherwise I don't understand what the problem asks for.2012-11-20
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    Unless, or course, the problem asks for $\lim 2$, the limit of a constant sequence, which is of course $2$. But I doubt that this is the case.2012-11-20
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    @DanShved sorry sir. The question will be: Let f:[-1,1]\rightarrow \ R be continuous.Assume that \int_{-1}^{1}f(t)dt =2.Then lim \int_{-1}^{1}f(t)sin^2(nt)dt equals to (a)0 (b)1 (c)f(1)-f(-1)(d)does not exist.2012-11-20
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    @budha: Did you learn [Riemann-Lebesgue Lemma](http://mathworld.wolfram.com/Riemann-LebesgueLemma.html)?2012-11-20
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    For a starting, the constant $f=1$ is an easy choice, and if the statement is true in this form, (i.e. doesn't depend too much on $f$), then it must give the right result.2012-11-20
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    @Pambos No sir, i did not learn it.2012-11-20
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    Then, what are the titles of the recent lectures/sections in the book?2012-11-20
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    @Berci Sir, i came across the problem in an exam paper.2012-11-20
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    Then you did the best. I personally dislike these quizes with given possible answers, but if that happens, just pick the easiest example. Anyway, the Riemann-Lebesgue lemma could help you, write $\sin^2$ as $1-\cos^2$ then use $2\cos^2x-1=\cos(2x)$..2012-11-20
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    @budha: In the exam you didn't have to justify your answer?2012-11-20
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    @Pambos You are right ,sir.But i did not like the way i solved it.So i wanted to know the theorem which would be useful to solve it.2012-11-20
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    @budha: In my opinion the best approach is what Berci suggested but uses a theorem (lemma) you didn't learn. In what course did you take the exams?2012-11-20
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    @Pambos I have used the formula as suggested by Berci. \frac{1}{2}\int_{-1}^{1}f(t)(1-cos (2nt))dt=\frac{1}{2}\int_{-1}^{1}f(t)dt-\frac{1}{2}\int_{-1}^{1}f(t)dtcos(2nt)dt=\frac{1}{2}\ast 2-0=12012-11-20

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Write $\sin^2(nx)=\dfrac{1-\cos{(2nx)}}{2}$.
Then $$\displaystyle{\int_{-1}^{1}f(t)\sin^2(nt)\,dt=\frac{1}{2}\int_{-1}^{1}f(t)\,dt-\frac{1}{2}\int_{-1}^{1}f(t)\cos(2nt)\,dt=1-\frac{1}{2}\int_{-1}^{1}f(t)\cos(2nt)\,dt}.$$ From Riemann-Lebesgue Lemma ( proof )we have $$\lim_{n\to\infty} \int_{-1}^{1}f(t)\cos(2nt)\,dt=0.$$ Therefore $\displaystyle{\lim_{n\to\infty} \int_{-1}^{1}f(t)\sin^2(nt)\,dt=1.}$