Posted by John on Thursday, March 7, 2013 at 12:03am.
Put x = 3 sinh(t), then:
Integral from 0 to 4 of
x^3 sqrt{9+x^2} dx =
Integral from 0 to arcsinh(4) of
3^5 sinh^3(t) cosh^2(t) dt =
3^5 Integral from 0 to arcsinh(4) of
sinh(t) [cosh^2(t) - 1] cosh^2(t) dt =
3^5 Integral from 0 to arcsinh(4) of
[cosh^2(t) - 1] cosh^2(t) dcosh(t)
Putting cosh(t) = u gives:
3^5 Integral from 1 to sqrt(17) of
[u^2 - 1] u^2 du =
3^5 {sqrt(17)[1/5 17^2 - 1/3 17] + 2/15 }
You then need to round this off to below.
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