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How to evaluate the integral $\displaystyle\int_0^r x^2\cos x\,dx$ for $r\in\mathbb{R}$ without using integration by parts?

And the hint is differentiate $\displaystyle\int_0^r\cos(tx)\,dx$ twice with respect to $t$.

The hint does not help. Can someone help me?

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    Well, have you done what the hint tell you to?2012-10-16
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    Yes. That's why I am asking for help2012-10-16
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    You might want to take a look at [this](http://en.wikipedia.org/wiki/Leibniz_integral_rule).2012-10-16
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    You might want to put the result you got into the question then. The clearer you make where your problem lies the higher the chance that the answers will be usefull to you.2012-10-16
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    The second integral is easy to evaluate. Call it $G(t)$. Differentiation under the integral sign shows that the integral you are after is closely related to $G''(1)$.2012-10-16
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    @EuYu this helps a lot but we haven't proven this rule. But thanks a lot. I'll think about it.2012-10-16
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    @FrankLu: What you noted as a hint was exactly the same as Eu Yu did below (+1).2012-10-16
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    @BabakSorouh exactly. The difficulty is I haven't learnt about differentiation under integral sign. That's what I have to think about.2012-10-16
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    Possible duplicate of [Integrating $x^2e^{-x}$ using Feynman's trick?](https://math.stackexchange.com/questions/943141/integrating-x2e-x-using-feynmans-trick)2018-04-02

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Let us write $$I(t) = \int_0^r \cos(tx)\ dx$$ Then from the Leibniz rule, we have $$\frac{d^2}{dt^2}I(t) = \int_0^r\frac{\partial^2}{\partial t^2}\cos(tx)\ dx=\int_0^r x^2\cos(tx)\ dx$$ Therefore your original integral is given by $$\frac{d^2}{dt^2}I(t)\Bigg|_{t=1}$$ Can you take it from here?

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    Thanks. I will try to prove this rule before using this2012-10-16