# Integrals with u substitution

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

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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

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 $$\underbrace{\int\frac{1}{\sqrt{1-\sin^2(t)}}\color{red}{\cos(t)dt}=\require{cancel}\int \frac{1}{\cancel{\cos(t)}}\cancel{\cos(t)}dt}_{\text{using the identity} \quad\sin^2(t)+\cos^2(t) \equiv 1}=\int1dt=\color{green}{t}+C=\boxed{\color{green}{\arcsin(u)+C}}.$$

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$$\frac{d}{dx}\sin^{-1}(u)=\frac1{\sqrt{1-u^2}}\frac{du}{dx}$$

And therefore the integral would be the reverse of that.

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\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.

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