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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 at 0:12
    
got it thanks ! –  Mark Feb 12 at 0:14
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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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