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It is said at the last discussion (The link toward the discussion) about integration by substitution $f$ must be bijective over $[a,b]$, with this example to illustrate :

$\int_{-1}^1 x^2\mathrm{d}x$ with $u =x^2$

I have found one example which $f$ is not bijective over $[a,b]$ and the u-substitution is suitable.

$\displaystyle I=\int_{-\pi/2}^{\pi/2} \cos(x)\cos(\sin(x))dx$

So if $\varphi(x) =\sin(x)\implies \displaystyle I=\int_{\varphi(-\pi/2)}^{\varphi(-\pi/2)} \varphi'(x)\cos\left(\varphi(x)\right)dx\iff\int_{-1}^1\cos(u)du=2\sin(1)$

And my question is :

The condition on the function $cos$ is : $\cos\in \mathcal{C}^1([\varphi(a),\varphi(b)],\mathbb{R})$ Am I right ?

And for $\int_{-1}^1 x^2\mathrm{d}x$ with $u =x^2$

$\implies \displaystyle \int_{1}^1 \dfrac{\sqrt{u}}{2} du$ and I don't think $f$ can be continuous over one point!?

Am I right?

Stu
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    I'm a bit confused on what you're asking for the first part. $x\to\sin(x)$ is a bijective mapping on the interval in question – TomGrubb Aug 03 '17 at 04:02
  • $f:x\to \cos x \cos (\sin x)$ is not bijective – Stu Aug 03 '17 at 04:07
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    Related: https://math.stackexchange.com/questions/801307/does-the-change-of-variable-function-have-to-be-injective – symplectomorphic Aug 03 '17 at 04:10
  • @ThomasGrubb, Yes it's right, I found a bad example. I will find an another example. But for $\int_{-1}^1 x^2\mathrm{d}x$ does it make sense my reasoning? – Stu Aug 03 '17 at 04:21
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    @stu $x\mapsto \cos x\cos (\sin x)$ is not the function used for the subsitution. $x\mapsto \sin x $ is, and it is bijective over $[-\pi/2;\pi/2)$ – Graham Kemp Aug 03 '17 at 04:21
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    If $u=x^2$, then $du=2xdx$. But $x=\sqrt{u}$ when $x>0$ and $x=-\sqrt{u}$ when $x<0$. So $$\int_{-1}^1x^2,dx=\int_{-1}^0x^2dx+\int_0^1x^2dx=0.5\int_1^0(-\sqrt{u}),du+0.5\int_0^1\sqrt{u},du$$ But the first term gets flipped to $0.5\int_0^1\sqrt{u},du$, so the sum is $\int_0^1\sqrt{u},du$, which is indeed equal to $\int_{-1}^1x^2,dx$. – symplectomorphic Aug 03 '17 at 04:24
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    The point is that you carried out the substitution incorrectly. It is not true that $x=\sqrt{u}$ for all $x$ in $(-1,1)$. – symplectomorphic Aug 03 '17 at 04:27
  • @symplectomorphic Thanks, I was looking for an argument because $\varphi$ can be bijective. Good night!! – Stu Aug 03 '17 at 04:39

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