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 somehow lose the inverse sine portion of the answer.

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

  • $\begingroup$ 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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.