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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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    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