Show that: `int_0^(pi//2)f(sin2x)sinxdx=sqrt(2)int_0^(pi//4)f(cos2x)cosxdxdot`

Explanation

Let `I=int_(0)^(pi//2)f(sin2x)sinx dx`
`=int_(0)^(pi//2)f{"sin"2(1/2pi-x)}sin(1/2 pi-x)dx`
`:. I=int_(0)^(pi//2)f(sin2x)cosx dx`
Then adding 1 and 2 we have
`2I=int_(0)^(pi//2)f(sin2x)(sinx+cosx)dx`
`=sqrt(2)int_(0)^(pi//2)f(sin2x)sin(x+(pi)/4)dx`
To get `f(cos 2x)` on RHS, we have to substitute `(pi)/2-2theta =2x`.
`:. dx=d theta`
Also, when `x=0, theta pi//4`. and when `x=pi//2, theta=-pi//4`.
`:. 2I=sqrt(2)int_(-pi//4)^(pi//4)f(cos 2theta) cos theta d theta `
`=2sqrtint_(0)^(pi//4) f(cos 2 theta) cos theta d theta`
[as `g(theta)=f(cos 2theta) cos theta` is an even function)
`:.I=sqrt(2)int_(0)^(pi//4)f(cos2x)cosx dx`