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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.

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  • $\begingroup$ Hint: Let $u = sinx$ and $du = cosx dx$. $\endgroup$
    – T. Fo
    Jan 8, 2019 at 17:18
  • $\begingroup$ 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... $\endgroup$ Jan 8, 2019 at 17:18
  • $\begingroup$ I believe one usually uses trig substitution for such things, try $u = \sin x$ $\endgroup$
    – gt6989b
    Jan 8, 2019 at 17:18
  • $\begingroup$ Oh I misread, my hint is for substitution. I'm not sure about IbP. $\endgroup$
    – T. Fo
    Jan 8, 2019 at 17:19
  • $\begingroup$ @PeterForeman Thanks I did it the other way already thanks though, I looked up another similar question. $\endgroup$ Jan 8, 2019 at 17:19

1 Answer 1

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

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