Need help with Integration problem?

Question

integral {(x^2)/ [sq root of (4-x^2)]} dx

Answer 1

Integral [ (x^2 / sqrt(4 - x^2)) dx ]

This has to be solved using trigonometric substitution.

Let x = 2sin(t). Then
dx = 2cos(t) dt, and our substitution becomes

Integral [ {4sin^2(t) / sqrt(4 - 4sin^2(t))} 2cos(t) dt ]

Skipping all the details, this reduces to

4 * Integral [ sin^2(t) ] dt

Which we use the half angle identity for

4 * Integral [ (1/2) (1 - cos(2t))] dt

2 * Integral [ 1 - cos(2t) ]

and this integrates easily now, as

2 * (t - (1/2) sin(2t)) + C, or

2t - sin(2t) + C

Using the double angle identity, we get

2t - 2sin(t)cos(t) + C

Now, since x = 2sin(t), then sin(t) = x/2 = opp/hyp

Since opp = x and hyp = 2, it follows that, by Pythagoras,
adj = sqrt(4 - x^2). This means

sin(t) = opp/hyp = x/2
cos(t) = adj/hyp = sqrt(4 - x^2)/2

Also, since sin(t) = x/2, t = arcsin(x/2). So

2t - 2sin(t)cos(t) + C

becomes

2arcsin(x/2) - 2[x/2] [sqrt(4 - x^2)/2] + C

2arcsin(x/2) - (1/2) x sqrt(4 - x^2) + C

Answer 2

It's a trigonometric substitution.

Note that when I type o, assume I mean theta, or the Greek letter of your choice,

As it's in the form sqr(a^2-x^2), we would make x=asino, or 2sino, and we must chain rule it, therefore:

integral of: (sino)^2/sqr(4-4(sino)^2) * coso
simplify the denominator to sqr(4(1-(sino)^2)) or sqr(4(coso)^2) or 2coso. So now we have:
integral of (sin2o)^2/2 (the cosines cancel out)
pull the 1/2 out front:
1/2 * integral of (sino)^2
so
1/2 * (o/2 -sin2o/4)
or
o/4 - sin2o/8
however, now we need to go back from o's into x's, so:
x=2sino
and o = arcsin(x/2). This would make your final answer(if I'm not mistaken):

(arcsin(x/2)/4)-(x/8) + C

Answer 3

Your integral has the square root of a quadratic polynomial. Whenever you have the square root of a polynomial use trigonometric substitution. In particular, factor two out of the denominator of the integrand to get

x^2/(2 sqrt(1-(x/2)^2))

make the substitution u = x /2, and then you should have one of the normal forms for trig sub.

Answer 4

Answer: (1/2) { -x [sq rt (4 - x^2)] + 4 [arcsin(x/2)] } + C

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