`int_(0)^(2pi)cos^(5)x dx`

Explanation

`int_(0)^(2pi)cos^(5)x dx= int_(0)^(pi)cos^(5)x dx`
`[because int_(0)^(a)f(x)=2 int_(0)^(a)f(x)dx," जहाँ "f(2a-x)=f(x), because cos^(5)(2pi-x)=cos^()x]`
`=2xx0=0`
`(because int_(0)^(2a)f(x)=0" यदि "f(2a-x)=-f(x), " यहाँ "2a=pi therefore cos^(5)(pi-x)=-cos^(5)x)`