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To compute the following integral

$$\int_{-1}^{1} \sqrt{1-x^2} \, dx$$

we use trig substitution and introduce the change of variable $x := \sin(\theta)$ where

$$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$$

To compute the differential $dx$ with respect to $\theta$, we let

$$\frac{dx}{d\theta} = \cos(\theta) \iff dx = \cos(\theta) \, d\theta$$

and we eventually have

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin(\theta)^2} \cos(\theta) \, d\theta$$

However, to get $dx$ with respect to $\theta$, we cancel out the differentials by separating them in the last step which does not seem very formal from a mathematical standpoint. What is the formal basis behind the substitution of $dx$ by $\cos(\theta) \, d\theta$ apart from this algebraic manipulation?

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    $\begingroup$ See THIS. $\endgroup$
    – Mark Viola
    Oct 18 '20 at 1:22
  • $\begingroup$ @MarkViola in the second example with the trig substitution, they do not explain how $dx$ is derived. I need help with the following line: "The substitution $x=\sin u$ implying $dx=\cos udu$ is useful because $\sqrt{1 - \sin^2 u} = \cos u$." $\endgroup$
    – explogx
    Oct 18 '20 at 1:33
  • $\begingroup$ Are you aware of the definition of a differential of a function? @eigenslacker $\endgroup$
    – RyanK
    Oct 18 '20 at 1:39
  • $\begingroup$ Read the sections entitled "Definite Integral" and "Proof." $\endgroup$
    – Mark Viola
    Oct 18 '20 at 2:01
  • $\begingroup$ @MarkViola thank you. Indeed the article on Wikipedia helped me a lot. $\endgroup$
    – explogx
    Oct 19 '20 at 9:48
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Got it. Trig substitution is $u$-sub applied backwards where $x$ is a function of $\theta$ so that

$$\int_{a}^{b} f(\varphi(\theta))\varphi'(\theta) \, d\theta = \int_{\varphi(a)}^{\varphi(b)} f(x) \, dx$$

where $x := \varphi(\theta)$ so by identification we also have $dx := \varphi'(\theta) \, d\theta$. Granted that $\varphi$ is continuous and differentiable on the bounds of integration.

Applying the same reasoning to the following integral

$$\int_{-1}^{1} \sqrt{1 - x^2} \, dx$$

we eventually obtain

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 - \sin(\theta)^2} \cos(\theta) \, d\theta = \int_{\sin\left(-\frac{\pi}{2}\right)}^{\sin\left(\frac{\pi}{2}\right)} \sqrt{1 - x^2} \, dx$$

where $x := \sin(\theta)$ and $dx := \cos(\theta) \, d\theta$.

This method is justified by the fact that $\sin(\theta)$ when

$$-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2}$$

is continuous and differentiable.

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