Tích phân - Trích đề thi thử ĐH (sưu tầm)

Đề bài: Tính tích phân : $I = \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi }{2}} x\sin x \left(5\sin x +x\cos x \right)\text{d}x.$
Giải:
Tích phân đã cho viết thành: $I = \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi }{2}}(5x\sin^2x+x^2\sin x\cos x)dx$

 $\ I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\left( {5x\frac{{1 - \cos 2x}}{2} + \frac{{{x^2}}}{2}\sin 2x} \right)} dx = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{5x}}{2}} dx - \frac{5}{2}\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} x \cos 2xdx + \frac{1}{2}\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {{x^2}} \sin 2xdx$
$\  = \frac{{5{x^2}}}{4}|_{\frac{\pi }{6}}^{\frac{\pi }{2}} - \frac{5}{2}{I_1} + \frac{1}{2}{I_2}.$

• Tính $I_2$

          Đặt $u=x^2\implies du=2xdx$ và $dv=\sin2xdx\implies v=-\dfrac{1}{2}\cos2x$
          $\displaystyle I_2=-\frac{x^2}{2}\cos2x\Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{2}}+I_1=\frac{19}{144}\pi ^2+I_1$

• Tính $I_1$

         Đặt $u=x\implies du=dx$ và $dv=\cos2xdx\implies v=\dfrac{1}{2}\sin2x$
         $I_1=\dfrac{x}{2}\sin2x\Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{2}}-\dfrac{1}{2}\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi }{2}}\sin2xdx=\frac{x}{2}sin2x\Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{2}}+{\frac{1}{4}}\cos2x\Bigg| _{\frac{\pi}{6}}^{\frac{\pi}{2}}=-\frac{\sqrt{3}{\pi}}{24}-\frac{3}{8}$

                                 Do đó $I=\dfrac{11}{32}\pi ^2+\dfrac{\sqrt{3}}{12}\pi +\dfrac{3}{4}$