a) $ \int\limits_{0}^{\pi/2} (2\sin x+3)\cos xdx= \int\limits_{0}^{\pi/2} 2\sin x cos xdx+ 3\int\limits_{0}^{\pi/2}\cos xdx$
$=\left[ {\sin^2 x +3\sin x} \right]_{0}^{\pi/2}=4$

b) Ta có
$2\sin6x.\sin2x=\cos 4x -\cos 8x$
Suy ra
$ \int\limits_{0}^{\pi/6} (\sin6x.\sin2x-6)dx= \left[ {-6x +\dfrac{1}{8}\sin 4x -\dfrac{1}{16}\sin 8x} \right]_{0}^{\pi/6}=\dfrac{3\sqrt 3}{32}-\pi$

a) Đặt $t=\cos x \Rightarrow dt =-\sin xdx$
$ \int\limits_{0}^{\pi/2} \sin^{5}xdx = \int\limits_{0}^{1} (1-t^2)dt=\left[ {t-\dfrac{2}{3}t^3+\dfrac{1}{5}t^5} \right]_{0}^{1} =\dfrac{8}{15}$

b)
Đặt $t= \dfrac{2\tan\dfrac{x}{2}+ 1}{\sqrt 3}\Rightarrow dt =\frac{1}{\sqrt 3}(1:\cos^2 \dfrac{x}{2})dx$
Bạn tự chứng minh đẳng thức sau xem như bài tập nhé.
$\dfrac{1}{2+ \sin x}=(1:\left[ {\left ( \dfrac{2\tan\dfrac{x}{2}+ 1}{\sqrt 3} \right )^2+1} \right]).\frac{2}{3}(1:\cos^2 \dfrac{x}{2})dx$
Suy ra
$\int\limits_{0}^{\pi/2}\dfrac{1}{2+ \sin x}dx=\int\limits_{\frac{1}{\sqrt 3}}^{\sqrt 3}\frac{2}{\sqrt 3}.\dfrac{1}{t^2+1}dt=\left[ {\frac{2}{\sqrt 3}\arctan t} \right]_{\frac{1}{\sqrt 3}}^{\sqrt 3}=\frac{2}{\sqrt 3} .\frac{\pi}{6}$

a) Ta có
$\sin^3 x \cos 5x =\sin^3 x \cos (4x+x) =-\sin^4 x \sin (4x)+\sin^3 x \cos x \cos 4x=\left ( \dfrac{1}{4} \sin^4 x\cos 4x\right )'$
Suy ra
$\int\limits_{0}^{\pi} \sin^{3}x\cos5xdx=\left[ { \dfrac{1}{4} \sin^4 x\cos 4x} \right]_{0}^{\pi}=0$