Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone please explain how the integration step highlighted in the red rectangle was worked out? enter image description here

share|cite|improve this question
Do you know what is the derivative of $\arcsin$ (AKA $\sin^{-1}$)? – Git Gud Aug 2 '14 at 23:49
Thank you solves my question – MacUser Aug 2 '14 at 23:50
My answer shows how to use the chain rule to show quickly that $\dfrac{d}{du}\sin^{-1} u = \dfrac{1}{\sqrt{1-u^2}}$. ${}\qquad{}$ – Michael Hardy Aug 3 '14 at 0:52
up vote 1 down vote accepted

Fundamentally, you're asking

Why is $$\int \frac{1}{\sqrt{1-u^2}}du=\arcsin(u)+C \quad ?$$

Let $\underbrace{u=\sin(t)}_{\iff \color{green}{t=\arcsin(u)}} \implies \frac{du}{dt}=\cos(t) \iff \color{red}{du=\cos(t)dt} $.

Then we've got $$\underbrace{\int\frac{1}{\sqrt{1-\sin^2(t)}}\color{red}{\cos(t)dt}=\require{cancel}\int \frac{1}{\cancel{\cos(t)}}\cancel{\cos(t)}dt}_{\text{using the identity} \quad\sin^2(t)+\cos^2(t) \equiv 1}=\int1dt=\color{green}{t}+C=\boxed{\color{green}{\arcsin(u)+C}}.$$

share|cite|improve this answer


And therefore the integral would be the reverse of that.

share|cite|improve this answer

$$ \begin{align} u & = \sin w \\[8pt] \frac{du}{dw} & = \cos w \tag 1 \\[8pt] \frac{dw}{du} & = \frac{1}{\cos w} = \frac{1}{\sqrt{1-\sin^2 w}} = \frac{1}{\sqrt{1-u^2}} \tag 2 \\[8pt] \frac{d}{du}\arcsin u & = \frac{1}{\sqrt{1-u^2}}. \\[8pt] \text{or if you prefer,} \\ \frac{d}{du} \sin^{-1} u & = \frac{1}{\sqrt{1-u^2}}. \end{align} $$

The step from $(1)$ to $(2)$ is just the chain rule.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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