Calculus
posted by John on .
Given that \displaystyle \int_0^4 x^3\sqrt{9+x^2} dx = a, what is the value of \lfloor a \rfloor?

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.