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

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?

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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

3 Answers 3

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!

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

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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

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

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