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

Answer 1

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.