下列计算结果正确的是()A. int_(0)^pi x cos x dx = -2B. int_(1)^2 x ln x dx = ln 4 - (3)/(4)C. int_(0)^((pi)/(2))^2 cos sqrt(x) dx = pi - 2D. int_((pi)/(4))^(pi)/(3) (x)/(sin^2 x) dx = ((1)/(4) - (sqrt(3))/(9) )pi

本题主要考查定积分的计算计算，解题思路是分别对每个选项中的定积分进行计算，然后与选项中给出的结果进行对比。

选项A

计算$\int_{0}^{\pi} x \cos x dx$，使用分部积分法$\int u dv = uv - \int v du$，令$u = x$，$dv = \cos x dx$，则$du = dx$，$v = \sin x$。
$\int_{0}^{\pi} x \cos x dx = [x \sin x]_{0}^{\pi} - \int_{0}^{\pi} \sin x dx$
$[x \sin x]_{0}^{\pi} = \pi \sin \pi - 0 \sin 0 = 0$
$\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = -\cos \pi + \cos 0 = -(-1) + 1 = 2$
所以$\int_{0}^{\pi} x \cos x dx = 0 - 2 = -2$，选项A正确。

选项B

计算$\int_{1}^{2} x \ln x dx$，使用分部积分法$\int u dv = uv - \int v du$，令$u = \ln x$，$dv = x dx$，则$du = \frac{1}{x} dx$，$v = \frac{1}{2} x^2$。
$\int_{1}^{2} x \ln x dx = [\frac{1}{2} x^2 \ln x]_{1}^{2} - \int_{1}^{2} \frac{1}{2} x^2 \cdot \frac{1}{x} dx$
$[\frac{1}{2} x^2 \ln x]_{1}^{2} = \frac{1}{2} \cdot 2^2 \ln 2 - \frac{1}{2} \cdot 1^2 \ln 1 = 2 \ln 2$
$\int_{1}^{2} \frac{1}{2} x^2 \cdot \frac{1}{x} dx = \frac{1}{2} \int_{1}^{2} x dx = \frac{1}{2} [\frac{1}{2} x^2]_{1}^{2} = \frac{1}{4} (2^2 - 1^2) = \frac{3}{4}$
所以$\int_{1}^{2} x \ln x dx = 2 \ln 2 - \frac{3}{4} = \ln 4 - \frac{3}{4}$，选项B正确。

选项C

计算$\int_{0}^{(\frac{\pi}{2})^2} \cos \sqrt{x} dx$，令$t = \sqrt{x}$，则$x = t^2$，$dx = 2t dt$。
当$x = 0$时，$t = 0$；当$x = (\frac{\pi}{2})^2$时，$t = \frac{\pi}{2}$。
$\int_{0}^{(\frac{\pi}{2})^2} \cos \sqrt{x} dx = \int_{0}^{\frac{\pi}{2}} \cos t \cdot 2t dt = 2 \int_{0}^{\frac{\pi}{2}} t \cos t dt$
使用分部积分法$\int u dv = uv - \int v du$，令$u = t$，$dv = \cos t dt$，则$du = dt$，$v = \sin t$。
$2 \int_{0}^{\frac{\pi}{2}} t \cos t dt = 2 [t \sin t]_{0}^{\frac{\pi}{2}} - 2 \int_{0}^{\frac{\pi}{2}} \sin t dt$
$[t \sin t]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \sin \frac{\pi}{2} - 0 \sin 0 = \frac{\pi}{2}$
$\int_{0}^{\frac{\pi}{2}} \sin t dt = [-\cos t]_{0}^{\frac{\pi}{2}} = -\cos \frac{\pi}{2} + \cos 0 = 1$
所以$\int_{0}^{(\frac{\pi}{2})^2} \cos \sqrt{x} dx = 2 \cdot \frac{\pi}{2} - 2 \cdot 1 = \pi - 2$，选项C正确。

选项D

计算$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} dx$，先对$\frac{x}{\sin^2 x}$进行变形，$\frac{x}{\sin^2 x} = x \csc^2 x$。
使用分部积分法$\int u dv = uv - \int v du$，令$u = x$，$dv = \csc^2 x dx$，则$du = dx$，$v = -\cot x$。
$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} dx = [-x \cot x]_{\frac{\pi}{4}}^{\frac{\pi}{3}} + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x dx$
$[-x \cot x]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -\frac{\pi}{3} \cot \frac{\pi}{3} + \frac{\pi}{4} \cot \frac{\pi}{4} = -\frac{\pi}{3} \cdot \frac{\sqrt{3}}{3} + \frac{\pi}{4} \cdot 1 = \frac{\pi}{4} - \frac{\sqrt{3} \pi}{9}$
$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x dx = [\ln |\sin x|]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \ln |\sin \frac{\pi}{3}| - \ln |\sin \frac{\pi}{4}| = \ln \frac{\sqrt{3}}{2} - \ln \frac{\sqrt{2}}{2} = \frac{1}{2} \ln \frac{3}{2}$
所以$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{x}{\sin^2 x} dx = \frac{\pi}{4} - \frac{\sqrt{3} \pi}{9} + \frac{1}{2} \ln \frac{3}{2} \neq (\frac{1}{4} - \frac{\sqrt{3}}{9}) \pi$，选项D错误。