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?

  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.