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

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.

• Hint: Let $u = sinx$ and $du = cosx dx$. Jan 8, 2019 at 17:18
• You forgot the sign in the integration by parts formula it should be a negative integral. I don't think you can solve this integral using integration by parts... Jan 8, 2019 at 17:18
• I believe one usually uses trig substitution for such things, try $u = \sin x$ Jan 8, 2019 at 17:18
• Oh I misread, my hint is for substitution. I'm not sure about IbP. Jan 8, 2019 at 17:19
• @PeterForeman Thanks I did it the other way already thanks though, I looked up another similar question. Jan 8, 2019 at 17:19

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