Tính tích phân $I = \int \limits_0^{\frac{\pi}{6}} x.\sin x\cos^2x dx$

Tính tích phân $I =\displaystyle \int \limits_0^{\frac{\pi}{6}} x.\sin x\cos^2x dx$

Giải

Đặt

$ u=x \Rightarrow du=dx \\ dv=\sin x\cos^2xdx =-\cos^2x d(\cos x) \Rightarrow v=-\dfrac{\cos^3 x}{3}$

$I=-\dfrac{x\cos^3 x}{3}\Big| _{0}^{\frac{\pi }{6}} + \dfrac{1}{3}\int_{0}^{\frac{\pi}{6}} \cos^3 x dx=-\dfrac{\pi\sqrt{3}}{48}+\dfrac{1}{3}I_1$

Ta có $I_1=\int_{0}^{\frac{\pi }{6}} (1-\sin^2 x)\cos x dx \\ =-\int_{0}^{\frac{\pi }{6}} (1-\sin^2 x) d(\sin x) \\ = \left(x-\dfrac{\sin^3 x}{3} \right)\Big| _{0}^{\frac{\pi }{6}}=\dfrac{\pi}{6}-\dfrac{1}{24}$

Vậy $I=-\dfrac{\pi\sqrt{3}}{48}+\dfrac{\pi}{18}-\dfrac{1}{72}$