IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - INTEGRATE THE FUNCTION- f(x)= xf(cosx) FROM 0 to 2 pie ?

man111

f(x)=_{[0 ]}

^{[2pi ]} xf(cosx)dx

put x=2pi-x in yhis integration.

f(2pi-x)=_{0 ]}^{[2pi ]} (2pi-x)f(cos(2pi-x))dx=2pi ₀^{[2pi ]}f(cosx)dx-_{0 ]}^{[2pi ]} xf(cosx)dx

f(x)=f(2pi-x)=2pi ₀^{[2pi ]}f(cosx)dx-f(x)

2f(x)=2pi ₀^{[2pi ]}f(cosx)dx=

f(x)=₀^{[2pi ]}f(cosx)dx=2₀^{[pi ]}f(cosx)dx=