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I was told by my maths teacher that the following procedure is wrong but didn't really understand why so I hope someone here can explain it to me.

We want to prove that $\int_0^{π/2} f(sinx)dx=\int_0^{π/2} f(cosx)dx$

I thought of setting $sinx=cosu$ a substitution which led to the desired result. Why is this wrong?

  • Do you know change of variable formula ? – Brian Ding Apr 23 '15 at 23:16
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    Your setting actually leads to a solution, as long as you correctly figure out $x$. Indeed, using the arc-cosine function we get $u=\arccos\sin x = \frac{\pi}{2}-x$, which yields the desired formula through change of variable. – Sangchul Lee Apr 23 '15 at 23:19
  • So you basically say that setting $u=π/2-x$ is the same thing? –  Apr 23 '15 at 23:22
  • What you are missing is substitution $\sin x \to \cos u$ doesn't guarantee $dx = du$. In fact, $du = -dx$ in this case. However, the final formula get saved by swapping the two limits. Next time, it will be much better/(safer for your teacher) to use the substitution $x \to u = \frac{\pi}{2} - x$. – achille hui Apr 23 '15 at 23:23

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There is nothing inherently wrong with that, except you have to be specific about where $u$ lies. Using $ \sin{x} = \cos{(\pi/2-x)} $, we have $$ \cos{(\pi/2-x)} = \cos{u}, $$ so we can choose $u=\pi/2-x$.

Going through with your substitution as-is requires some assumptions on the signs of the functions: $$ \cos{x} \, dx = -\sin{u} \, du, $$ and $$ \cos{x} = \sqrt{1-\sin^2{x}} = \sqrt{1-\cos^2{u}} = \sin{u}, $$ assuming that $\sin{u}>0$. Then the limits swap, and the above shows that actually $ dx = -du$, so you do find $$ \int_0^{\pi/2} f(\sin{x}) \, dx = -\int_{\pi/2}^{0} f(\cos{u}) \, du = \int_0^{\pi/2} f(\cos{u}) \, du $$

Chappers
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