Change variable in the integral doesn't work

Question

I need to show that for small angles $$f_N\left(\frac{\pi}{2N}\right) \approx \frac{2}{\pi}\int_0^\pi\frac{\sin(u)}{u}du$$ the only information that I have is that we have $$f_N(x) = \frac{2}{\pi}\int_0^x\frac{\sin(2Nt)}{\sin(t)}dt$$

My Attempt

So I tried witht he change of variable $2Nt = u$ so that $dt = \frac{du}{2N}$ and then substituted in $$f_N(x) = \frac{2}{\pi}\int_0^x\frac{\sin(u)}{\sin\left(\frac{u}{2N}\right)}\frac{du}{2N} \approx \frac{2}{\pi}\int_0^x\frac{\sin(u)}{u}du$$

But then when I plug in $x = \frac{\pi}{2N}$ I get $$f_N\left(\frac{\pi}{2N}\right) \approx \frac{2}{\pi}\int_0^{\frac{\pi}{2N}}\frac{\sin(u)}{u}du$$

What did I do wrong?

Answer 1

You didn't change the upper limit in the integral.

When $t=x$, $u = 2Nx$.

Then, when you put in $x = \frac{\pi}{2N}$, you get $u = \pi$.