BrandonS15
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Hello,
I’m computing the integral of 1/(4-x^2) and I’ve done the following:
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I tried checking my answer using symbolab however, It gave me a different result (a different method?) I tried graphing their solution compared with mine to see if its due to difference in constants and the functions don’t appear the same so I’ve done something wrong I’m assuming but I cannot see where my potential error lies.
Any help is appreciated, thanks in advance
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_gcx
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You've substituted back u = \dfrac 1 2 x not \sin u = \dfrac 1 2 x. Start by writing \sec u and \tan u in terms of \sin u. Everything before looks fine.

FWIW - partial fractions is easier here, writing 4 - x^2 = (2 - x)(2 + x).
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BrandonS15
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(Original post by _gcx)
You've substituted back u = \dfrac 1 2 x not \sin u = \dfrac 1 2 x. Start by writing \sec u and \tan u in terms of \sin u. Everything before looks fine.

FWIW - partial fractions is easier here, writing 4 - x^2 = (2 - x)(2 + x).
I see, simple mistake, however when I sketch the graph (red is the symbolab solution) (blue is mine) I see my solution is constrained between interval (-2,2), this would be due to the arcsin(x/2) so how would I turn my result into the red one?

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I see sec(arcsin(x/2)) is messy but I can’t seem to figure an identity linking sec and sin which I could apply to write this without using arcsin, how would I go about rewriting this more nicely and so it includes intervals (-inf,inf) like the red curve given by equation:
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_gcx
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(Original post by BrandonS15)
I see, simple mistake, however when I sketch the graph (red is the symbolab solution) (blue is mine) I see my solution is constrained between interval (-2,2), this would be due to the arcsin(x/2) so how would I turn my result into the red one?

Name:  826BA8C1-844E-4E77-A890-C7C0DAC1E559.jpg.jpeg
Views: 7
Size:  28.9 KB

I see sec(arcsin(x/2)) is messy but I can’t seem to figure an identity linking sec and sin which I could apply to write this without using arcsin, how would I go about rewriting this more nicely and so it includes intervals (-inf,inf) like the red curve given by equation:
Name:  8FC4AAF7-980A-4A01-A65C-505DDB0DA84C.jpg.jpeg
Views: 7
Size:  22.2 KB
By setting \sin u = \dfrac 1 2 x you've implicitly constrained -2 \le x \le 2 (but we obvs can't have x = \pm 2) because of the range of sine so you're right.

You can then rewrite this in the above form (consider \sin^2 u + \cos^2 u = 1 to get \cos u (up to a sign, but you can probably ignore that, the modulus makes it not matter anyway) and so \sec u and \tan u)

[But still you've technically only proved the integral for -2 < x < 2, this proof would not work to show for all x \ne \pm 2, for |x| > 2 you would have to use a substitution like \dfrac 1 2 x = \sec u or partial fractions. I think at A-level you can ignore this technicality. It'd be harder to overlook with a definite integral.]
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