So, I integrated this integral: $$I = \int\frac{dx}{(x^2+4)\sqrt{1-x^2}}$$ And got: $$I = \frac{1}{2\sqrt{5}}\arctan{\frac{x\sqrt{5}}{2\sqrt{1-x^2}}}+C$$ But, since $-1 < x < 1$, how does it affect the integral?
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The domain condition doesn't affect the indefinite integral. However, if you were to try using your result with a definite integral you'd need to make sure that the bounds satisfy your domain conditions. – Zach466920 Apr 30 '15 at 16:14
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I asked this because, in my workbook, it says the solution above is only for $0 < x < 1$ and for $-1 < x < 0$ it is: $$I=-\frac{1}{2\sqrt{5}}\arctan{\frac{x\sqrt{5}}{2\sqrt{1-x^2}}}+C$$ Why do we change the sign? – A6SE Apr 30 '15 at 16:17
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hey!, the integrate function is even! do you remember the relation between the derivate for $x<0$ and $x>0$? – L F Apr 30 '15 at 16:26
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I'm not sure, please explain. I'm really confused with even/odd thing and what derivative tells us. – A6SE Apr 30 '15 at 16:30
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So, our function is even and the returns same values for -x and x. So, how does it affect our restriction and why do we change sign, still confused :( – A6SE Apr 30 '15 at 16:33
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I actually did it very similarly to WA, like this: http://oi62.tinypic.com/utvuv.jpg and look what WA does in the last step, it really confuses me. – A6SE Apr 30 '15 at 17:23