# Formal basis for computing the differential in trig substitution

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?

• See THIS. Oct 18 '20 at 1:22
• @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 $\sqrt{1 - \sin^2 u} = \cos u$." Oct 18 '20 at 1:33
• Are you aware of the definition of a differential of a function? @eigenslacker Oct 18 '20 at 1:39
• Read the sections entitled "Definite Integral" and "Proof." Oct 18 '20 at 2:01
• @MarkViola thank you. Indeed the article on Wikipedia helped me a lot. Oct 19 '20 at 9:48

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.