Simple way to compute the integral $\int_{\sqrt{n\pi}}^{\sqrt{(n + 1)\pi}}\sin(x^2)x \mathrm dx= 1$

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?

• Substitution? $x^2 = t$. – Daniel Fischer Feb 6 '14 at 14:16
• something is missing in the integrant. – guest Feb 6 '14 at 14:18

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.
• I thought $\int{sin(x)} = 1 - cos(x)$. No? – Graduate Feb 6 '14 at 14:57
• That would be if you had $$\int_0^x \sin (t) \, dt,$$ but the indefinite integral is just $- \cos (x)$. – Mark Fantini Feb 6 '14 at 15:29
• 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. – Oscar Lanzi Jan 3 '17 at 13:29
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 $1$.