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It is not necessary to take any example of any integral, so I'll just drop it: $$u=\sin x; \ \ \ du=\cos x dx$$ which my question is rather here:

Is it possible to do: (squaring both sides) $$u^2=\sin^2(x)=1-\cos^2(x)$$ $$\cos(x)=\sqrt{1-u^2}$$ is it? I don't think it's true for the simple fact that $\sqrt{x^2}=|x|$ and not just $x$ but my question earlier today talked about how domain can be somewhat dealt with the constant $+C$ as shown in my question:

My question about different domains on indefinite integrals

So I'm very curious on whether it would be valid or not

Travis Willse
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Aley20
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    If we restrict to an interval where $\cos(x) \geq 0$, then there's no issue, but what would you propose happens on any subinterval where $\cos(x) < 0$? – Brian Moehring Dec 09 '22 at 08:01

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Of course squaring still yields a true identity, but as your comment regarding absolute value suggests (and as Brian Moehring made explicit in the comments) $\cos x = \sqrt{1 - u^2}$ cannot hold on any interval where $\cos x < 0$.

As an example, consider $$\int_0^\pi \sin x \cos^2 x \,dx .$$ The conventional substitution $v = \cos x$ quickly gives the value $\frac{2}{3}$.

On the other hand, taking $u = \sin x$, $du = \cos x\, dx$, and replacing $\cos x \rightsquigarrow \sqrt{1 - u^2}$ gives the incorrect value $$\int_0^0 u \sqrt{1 - u^2} \,du = 0 .$$

To avoid this error, we can first split our domain of integration into subintervals on which $\cos x \geq 0$ or $\cos x \leq 0$, say, $[0, \frac{\pi}{2}]$ and $[\frac{\pi}{2}, \pi]$, whereon we thus have $\cos x = \sqrt{1 - u^2}$ and $\cos x = -\sqrt{1 - u^2}$, respectively. Then, our original integral becomes \begin{align} \int_0^\pi \sin x \cos^2 x \,dx &= \int_0^\frac\pi2 \sin x \cos^2 x \,dx + \int_\frac\pi2^\pi \sin x \cos^2 x \,dx \\ &= \int_0^1 u \sqrt{1 - u^2} \,du + \int_1^0 u \left(-\sqrt{1 - u^2}\right) \,du \\ &= 2 \int_0^1 u \sqrt{1 - u^2} \,du \\ &=\frac23 . \end{align}

Travis Willse
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