# Thread: Help required with this definite integration equation!

1. ## Help required with this definite integration equation!

Integrate s/square root of (s squared + 4 squared) ds.

Definite integral has range 0 - 3

2. Originally Posted by Steve Mckenna
Integrate s/square root of (s squared + 4 squared) ds.

Definite integral has range 0 - 3
For $\displaystyle \int_0^3 \frac{s}{\sqrt{s^2+16}}\, ds$ try letting $\displaystyle u = s^2 +16.$

3. Originally Posted by Steve Mckenna
Integrate s/square root of (s squared + 4 squared) ds.

Definite integral has range 0 - 3
$\displaystyle \int {\frac{s} {{\sqrt {{s^2} + {4^2}} }}ds} = \frac{1} {2}\int {\frac{{2s}} {{\sqrt {{s^2} + 16} }}ds} =$$\displaystyle \left\{ \begin{gathered} \sqrt {{s^2} + 16} = t, \hfill \\ {s^2} = {t^2} - 16, \hfill \\ 2sds = 2tdt \hfill \\ \end{gathered} \right\} = \displaystyle = \frac{1} {2}\int {\frac{{2t}} {t}dt} = \int {dt} = t + C = \sqrt {{s^2} + 16} + C. \displaystyle \int\limits_0^3 {\frac{s} {{\sqrt {{s^2} + 16} }}ds = } \left. {\sqrt {{s^2} + 16} } \right|_0^3 = \sqrt {25} - \sqrt {16} = 5 - 4 = 1. 4. Originally Posted by DeMath \displaystyle \int {\frac{s} {{\sqrt {{s^2} + {4^2}} }}ds} = \frac{1} {2}\int {\frac{{2s}} {{\sqrt {{s^2} + 16} }}ds} =$$\displaystyle \left\{ \begin{gathered} \sqrt {{s^2} + 4} = t, \hfill \\ {s^2} = {t^2} - 16, \hfill \\ 2sds = 2tdt \hfill \\ \end{gathered} \right\} =$

$\displaystyle = \frac{1} {2}\int {\frac{{2t}} {t}dt} = \int {dt} = t + C = \sqrt {{s^2} + 16} + C.$

$\displaystyle \int\limits_0^3 {\frac{s} {{\sqrt {{s^2} + 16} }}ds = } \left. {\sqrt {{s^2} + 16} } \right|_0^3 = \sqrt {25} - \sqrt {16} = 5 - 4 = 1.$
Many thanks