If x satisfies the equation left( int_{0}^{1} | vedclass

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(D) Let $I_1 = \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt$. Since $f(t) = \frac{t^2 \sin 2t}{t^2 + 1}$ is an odd function (because $f(-t) = -f(t)$)

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$\pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$

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(D) Let $I_1 = \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt$. Since $f(t) = \frac{t^2 \sin 2t}{t^2 + 1}$ is an odd function (because $f(-t) = -f(t)$) and the interval $[-3, 3]$ is symmetric about $0$,$I_1 = 0$.Let $I_2 = \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1}$. We can write the denominator as $(t + \cos \alpha)^2 + \sin^2 \alpha$.Thus,$I_2 = \int_{0}^{1} \frac{dt}{(t + \cos \alpha)^2 + \sin^2 \alpha} = \left[ \frac{1}{\sin \alpha} 	an^{-1} \left( \frac{t + \cos \alpha}{\sin \alpha} ight) ight]_0^1$.$I_2 = \frac{1}{\sin \alpha} \left( 	an^{-1} \left( \frac{1 + \cos \alpha}{\sin \alpha} ight) - 	an^{-1} \left( \frac{\cos \alpha}{\sin \alpha} ight) ight)$.Using half-angle identities,$\frac{1 + \cos \alpha}{\sin \alpha} = \frac{2 \cos^2(\alpha/2)}{2 \sin(\alpha/2) \cos(\alpha/2)} = \cot(\alpha/2) = 	an(\pi/2 - \alpha/2)$.Also,$\frac{\cos \alpha}{\sin \alpha} = \cot \alpha = 	an(\pi/2 - \alpha)$.So,$I_2 = \frac{1}{\sin \alpha} (\pi/2 - \alpha/2 - (\pi/2 - \alpha)) = \frac{1}{\sin \alpha} (\alpha/2) = \frac{\alpha}{2 \sin \alpha}$.The equation becomes $\left( \frac{\alpha}{2 \sin \alpha} ight) x^2 - 0 \cdot x - 2 = 0$.$x^2 = \frac{4 \sin \alpha}{\alpha} \Rightarrow x = \pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$.

If $x$ satisfies the equation $\left( \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1} \right) x^2 - \left( \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt \right) x - 2 = 0$ for $0 < \alpha < \pi$,then the value of $x$ is

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