Tính tích phân sau: ∫sin^6 xdx/(sin^6 x+cos^6 x)

Question

Tính tích phân sau:

\(\int _{0}^{\frac{\pi }{2} }\frac{\sin ^{6} xdx}{\sin ^{6} x+cos^{6} x}\)

Answer 1

\(I=\int _{0}^{\frac{\pi }{2} }\frac{\sin ^{6} xdx}{\sin ^{6} x+cos^{6} x} \) 

 Đặt \(x=\frac{\pi }{2} -u\Rightarrow dx=-du\)

 Đổi cận: \(x=0\Rightarrow u=\frac{\pi }{2} ;x=\frac{\pi }{2} \Rightarrow u=0\)
\(\Rightarrow I=\int _{0}^{\frac{\pi }{2} }\frac{\sin ^{6} xdx}{\sin ^{6} x+cos^{6} x}  =\int _{0}^{\frac{\pi }{2} }\frac{cos^{6} udu}{\sin ^{6} u+cos^{6} u}  \)

\(\Rightarrow I=\int _{0}^{\frac{\pi }{2} }\frac{cos^{6} xdx}{\sin ^{6} x+cos^{6} x}\)  

\(\Rightarrow 2I=\int _{0}^{\frac{\pi }{2} }\frac{\sin ^{6} xdx}{\sin ^{6} x+cos^{6} x} +\int _{0}^{\frac{\pi }{2} }\frac{cos^{6} xdx}{\sin ^{6} x+cos^{6} x} \) 

\(=\int _{0}^{\frac{\pi }{2} }dx=\frac{\pi }{2}  \) \({\Rightarrow I=\frac{\pi }{4} }\)