If the function `f : [-1,1] to R` is continuous and even, then show that `int_(0)^(pi//2)f(cos2x)cosxdx=sqrt(2)int_(0)^(pi//4)f(sin2x)cosxdx`.

Explanation

`l=int_(0)^(pi//2)f(cos2x)cosxdx=int_(0)^(pi//2)f(cos{2((pi)/(2)-x)})cos((pi)/(2)-x)dx`
`=int_(0)^(pi//2)f(cos2x)sinxdx`, (`f` is an even function)
Hence, `2I=int_(0)^(pi//2)f(cos2x)(sinx+cosx)dx`
`=sqrt(2)int_(0)^(pi//2)f(cos2x)cos(x-(pi)/(4))dx`
`=sqrt(2)int_(-pi//4)^(pi//4)f(cos{2(u+(pi)/(4))})cosudu`, (setting `x=u+(pi)/(4)`)
`=sqrt(2)int_(-pi//4)^(pi//4)f(-sin2u)cosudu=sqrt(2)int_(-pi//4)^(pi//4)f(sin2u)cosudu` as `f` is even
`=2sqrt(2)int_(0)^(pi//4)f(sin2u)cosuduimpliesI=sqrt(2)int_(0)^(pi//4)f(sin2u)cosudu`
`=sqrt(2)int_(0)^((pi)/(4))f(sin2x)cosxdx`