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Is there a simple way to show that $$\int_{\sqrt{n\pi}}^{\sqrt{(n + 1)\pi}}\sin(x^2)x \mathrm dx= 1$$ if $n$ is even. We don't know how to integrate a multiple of functions ($\int{f(x)g(x)}$), but know how integrate $\sin(x)$ and $\cos(x)$. Is there some playing with trigonometric identities to show this?

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    $\begingroup$ Substitution? $x^2 = t$. $\endgroup$ – Daniel Fischer Feb 6 '14 at 14:16
  • $\begingroup$ something is missing in the integrant. $\endgroup$ – guest Feb 6 '14 at 14:18
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As suggested in the comments, $$ x\, dx = \frac{1}{2} d(x^2). $$ Hence $$ \int \sin (x^2) x \, dx = \frac{1}{2}\int \sin (x^2) \, d(x^2) = -\frac{1}{2}\cos (x^2) +C. $$ Therefore $$ \int_{(\sqrt{n \pi},\sqrt{(n+1)\pi})} \sin (x^2) x \, dx = -\frac{1}{2} \left( \cos ((n+1)\pi) - \cos (n\pi) \right), $$ and you conclude easily because $n$ is even.

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  • $\begingroup$ I thought $\int{sin(x)} = 1 - cos(x)$. No? $\endgroup$ – Graduate Feb 6 '14 at 14:57
  • $\begingroup$ That would be if you had $$\int_0^x \sin (t) \, dt,$$ but the indefinite integral is just $- \cos (x)$. $\endgroup$ – Mark Fantini Feb 6 '14 at 15:29
  • $\begingroup$ Actually the indefinite integral could be $1-\cos(x)$ if you wanted it to be. Arbitrary constant. One might write $-\cos(x)+C$ to cover all cases. $\endgroup$ – Oscar Lanzi Jan 3 '17 at 13:29
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As already commented, take $ t = x ^ 2 \implies \frac{1}{2} dt = x dx $. Therefore, the limits change to $ n \pi $ to $ (n + 1) \pi $. And the integral becomes:

$$ \int_{n \pi}^{(n + 1) \pi}{\sin{(t)} \frac{1}{2} dt} = {-\dfrac{1}{2} \left[ \cos{(t)}\right]_{n \pi}^{(n + 1) \pi}} = \dfrac{-1}{2} \left( (-1)^{2k + 1} - (-1)^{2k} \right) \tag{ Take $n = 2k$} $$

From there, you get the integration to be $ 1 $.

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