Evaluate this integral without using integration by parts

Question

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?

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. — fgp, Oct 16 '12 at 19:03
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 $G''(1)$. — André Nicolas, Oct 16 '12 at 19:05
@EuYu this helps a lot but we haven't proven this rule. But thanks a lot. I'll think about it. — Frank Lu, Oct 16 '12 at 19:07
@FrankLu: What you noted as a hint was exactly the same as Eu Yu did below (+1). — Mikasa, Oct 16 '12 at 19:09
@BabakSorouh exactly. The difficulty is I haven't learnt about differentiation under integral sign. That's what I have to think about. — Frank Lu, Oct 16 '12 at 19:14
Possible duplicate of Integrating $x^2e^{-x}$ using Feynman's trick? — BCLC, Apr 02 '18 at 20:27

score 4 · Accepted Answer · answered Oct 16 '12 at 19:06

4

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?

answered Oct 16 '12 at 19:06

EuYu

41,421

Thanks. I will try to prove this rule before using this – Frank Lu Oct 16 '12 at 19:11

Evaluate this integral without using integration by parts

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