Integration by Parts only of $\sqrt{1-u^2}$

Question

I am trying to integrate the function: $$f(x)=\sqrt{1-u^2}$$ I was using integration by parts to attack the problem, and it was: $$\int\sqrt{1-u^2}du$$ I set $g=\sqrt{1-u^2}$ and $dv=du$

Thus leading me to get: $$u\sqrt{1-u^2}+\int\frac{u^2}{\sqrt{1-u^2}}$$

from there I set $g=u$, and $dv=\frac{u}{\sqrt{1-u^2}}du$

$$I=u\sqrt{1-u^2}-u\sqrt{1-u^2}-I$$

I somehow lose the inverse sine portion of the answer.

Just using integration by parts is there a way I can get the right answer.

Answer 1

Let $I = \int\sqrt{1-u^2}\, du$.

\begin{eqnarray*} I &=& u\sqrt{1-u^2} + \int \frac{u^2}{\sqrt{1-u^2}} \, du \\ &=& u\sqrt{1-u^2} - \int \frac{1-u^2 - 1}{\sqrt{1-u^2}} \, du \\ &=& u\sqrt{1-u^2} - I +\int \frac{1}{\sqrt{1-u^2}} \, du \\ &=& u\sqrt{1-u^2} - I + \arcsin(u)\\ \end{eqnarray*}

It follows: $$2I = u\sqrt{1-u^2} + \arcsin(u) \leftrightarrow I = \frac{1}{2}\left( u\sqrt{1-u^2} + \arcsin(u)\right) (+ C)$$