Integrating $$1-y^2$$: Simplifying Arcsins

[NotaMathPerson]

$$\int(1-y^2)^\frac{1}{2}\,dy$$

I did trig substitution

$$y=\sin\theta$$
$$dy=\cos\theta\,d\theta$$

$$\int(1+\cos2\theta)d\theta$$
$$\arcsin\,y+\frac{1}{2}\sin(2\arcsin\,y)+c$$

How do I get rid of the arcsins?

 

[MarkFL]

After your trig. substitution and appliction of the double-angle identity for cosine, you should have:

\(\displaystyle I=\frac{1}{2}\int 1+\cos(2\theta)\,d\theta=\frac{1}{2}\left(\theta+\frac{1}{2}\sin(2\theta)\right)+C\)

Okay, now at this point we can apply a double-angle identity for sine and state:

\(\displaystyle I=\frac{1}{2}\left(\theta+\sin(\theta)\cos(\theta)\right)+C\)

Okay, now you can back-substitute for $\theta$, observing that:

\(\displaystyle \theta=\arcsin(y)\)

\(\displaystyle \sin(\theta)=y\)

\(\displaystyle \cos(\theta)=\sqrt{1-y^2}\)

And so we find:

\(\displaystyle I=\frac{1}{2}\left(\arcsin(y)+y\sqrt{1-y^2}\right)+C\)

And that's the final answer...no need to "get rid" of the inverse sine function. :D