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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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    @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

1 Answers 1

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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 $\displaystyle{\lim_{n\to\infty} \int_{-1}^{1}f(t)\sin^2(nt)\,dt=1.}$