Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\int_0^\frac{1}{2} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx $$

My question for this is can I take out the square root so it would be

$$\frac{1}{\sqrt{1-x^2}} \int_{0}^\frac{1}{2}{\sin^{-1}x} \,dx $$

So all i have to do is find the anti-derivative of $\sin^{-1}x$ then multiply the $\dfrac{1}{\sqrt{1-x^2}}$ back in afterwards? Is there a simpler way?

share|cite|improve this question
You can only take out constant factors from integrals. Otherwise you'd get silly stuff like $$\int_0^1 x\, dx = x\int_0^1 1\, dx = x \cdot 1 = x$$ which is certainly wrong because definite integrals are constant! – Clive Newstead Feb 12 '14 at 0:12
got it thanks ! – Mark Feb 12 '14 at 0:14
up vote 1 down vote accepted

Hint: let $u = \arcsin x$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$. You're not allowed to move the $\frac{1}{\sqrt{1-x^2}}$ term out in front since it has an $x$ in it and the variable of integration is $x$--so the integral depends on that term too!

share|cite|improve this answer

Hint: Use the fact that $\arcsin'x=\dfrac1{\sqrt{1-x^2}}$ .

share|cite|improve this answer

I may suggest that this integral must be like this Integral arc sin x d(arc sin x) from 0 to 1/2. thus the answer must be π^2/72

share|cite|improve this answer
Can you clarify what you mean? – user103828 Apr 1 '15 at 6:41
This has already been (at least as well) explained more than one year ago. – Did Apr 1 '15 at 14:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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