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

 A: $$\frac{d}{dx}\sin^{-1}(u)=\frac1{\sqrt{1-u^2}}\frac{du}{dx}$$
And therefore the integral would be the reverse of that.
A: 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 $$\underbrace{\int\frac{1}{\sqrt{1-\sin^2(t)}}\color{red}{\cos(t)dt}=\require{cancel}\int \frac{1}{\cancel{\cos(t)}}\cancel{\cos(t)}dt}_{\text{using the identity} \quad\sin^2(t)+\cos^2(t) \equiv 1}=\int1dt=\color{green}{t}+C=\boxed{\color{green}{\arcsin(u)+C}}.$$
A: $$
\begin{align}
u & = \sin w \\[8pt]
\frac{du}{dw} & = \cos w \tag 1 \\[8pt]
\frac{dw}{du} & = \frac{1}{\cos w} = \frac{1}{\sqrt{1-\sin^2 w}} = \frac{1}{\sqrt{1-u^2}} \tag 2 \\[8pt]
\frac{d}{du}\arcsin u & = \frac{1}{\sqrt{1-u^2}}. \\[8pt]
\text{or if you prefer,} \\
\frac{d}{du} \sin^{-1} u & = \frac{1}{\sqrt{1-u^2}}.
\end{align}
$$
The step from $(1)$ to $(2)$ is just the chain rule.
