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April 19, 2014

April 19, 2014

Posted by **John** on Thursday, March 7, 2013 at 12:03am.

- Calculus -
**Count Iblis**, Thursday, March 7, 2013 at 11:55amPut 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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