Q:

Integrals with u substitution

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

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

