Wish
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Q: \int \frac{1}{\sqrt{1-4x^2}} \ \mathrm{d}x

I got \mathrm{arcsin}(2x) but the answer states that it is \frac12\mathrm{arcsin}(2x). Where did the \frac12 come from?
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IChem
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(Original post by Wish)
Q: \int \frac{1}{\sqrt{1-4x^2}} \ \mathrm{d}x

I got \mathrm{arcsin}(2x) but the answer states that it is \frac12\mathrm{arcsin}(2x). Where did the \frac12 come from?
Dividing by the derivative of the inner function. (2x)
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Audrey Hepburn
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you have to say that 2x=u

then dx/du = 1/2

so you intergrate: 0.5[1/(1-u^2)]du

and you end up with the right answer
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saklut
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to integrate it you use root(a^2 - x^2). In this case x^2 is 4x^2 and hence, x is 2x, so you have to divide the integral by its derivate which is 2.
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Wish
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Is that always the case? For all the ones provided in the formula book. If it has a constant cooefficient then we have to have a fraction on the square root at the front.
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JSG1998
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This is how i do it, maybe you could find it useful...
you take the 4 out of the root i.e. 1 / sqrt(4 (1/4)+ x^2) which becomes 1/(2 sqrt( (1/4)+(x^2) )... then take the 1/2 out since its a constant and then integrate 1 / sqrt ( (1/4) + (x^2) ) and multiply the (1/2) at the end...
doing it this way always works for me
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