How do I get this function to a certain integral?

Question

I want ∫f(x)=1 on 0 to 1. what to I have to do to Sin(πx) to get that?

Answer 1

Evaluate the integral between the limits and get a value. Then divide sin(πx) by that value.

∫sin(πx)dx = -(1/π)cos(πx)

At x = 1, -(1/π)cos(πx) = 1/π
At x = 0 -(1/π)cos(πx) = -1/π

Subtract them to get 2/π. Divide sin(πx) by 2/π to get

(π/2)sin(πx)

Answer 2

Usually the integral of (sin x) from 0 to pi is 2.
The integral of (sin (pi x) ) from 0 to 1 is 2/pi (we kept same height but changes the base from pi to 1 (divided base by pi)
So to get this integral to be = 1, we have to start with 2/pi.. what times 2/pi = 1??? ===> we have to multiply then by (pi/2)....

So the answer is (pi/2)sin(pi/x)