Tính tích phân sau: ∫(3√cos x −3√sin x)dx

Question

Tính tích phân sau:

\( \int _{0}^{\frac{\pi }{2} }\left(\sqrt[{3}]{\cos x} -\sqrt[{3}]{\sin x} \right){\rm d}x\)

Answer 1

\(\int _{0}^{\frac{\pi }{2} }\left(\sqrt[{3}]{\cos x} -\sqrt[{3}]{\sin x} \right){\rm d}x      \)

 Đặt \(t=\frac{\pi }{2} -x\Rightarrow {\rm d}t=-{\rm d}x.\)  

Đổi cận: \(x=\frac{\pi }{2} \Rightarrow t=0;\, x=0\Rightarrow t=\frac{\pi }{2} \)

 Khi đó:  \(I=-\int _{\frac{\pi }{2} }^{0}\left(\sqrt[{3}]{\cos \left(\frac{\pi }{2} -t\right)} -\sqrt[{3}]{\sin \left(\frac{\pi }{2} -t\right)} \right){\rm d}t \)

\(=\int _{0}^{\frac{\pi }{2} }\left(\sqrt[{3}]{\sin t} -\sqrt[{3}]{\cos t} \right){\rm d}t=-I \Rightarrow 2I=0\)

 Vậy \(I=\int _{0}^{\frac{\pi }{2} }\left(\sqrt[{3}]{\cos x} -\sqrt[{3}]{\sin x} \right){\rm d}x =0\)

 Chú ý:  Ta có \(\int _{0}^{\frac{\pi }{2} }\sqrt[{3}]{\cos x} {\rm d}x= \int _{0}^{\frac{\pi }{2} }\sqrt[{3}]{\sin x} {\rm d}x \Rightarrow I=0\)