Hi everyone, just need help with this problem if anyone can point me in the right direction.
Evaluate \int {0}^{3} s/sqrt(s^2+4^2) ds
Thanks.
$\displaystyle \int\limits_0^3 {\frac{s}{\sqrt{s^2+16}}\,ds} = \left\{ \begin{gathered}s^2 = t^2- 16,\,s\,ds = t\,dt \hfill \\
0 \leqslant s \leqslant 3,\quad 4 \leqslant t \leqslant 5 \hfill \\
\end{gathered} \right\} =$
$\displaystyle = \int\limits_4^{5} {\frac{t}{{\sqrt {{t^2} - 16 + 16} }}\,dt} = \int\limits_4^5 {dt} =\, \Bigr. t \Bigr|_4^5 = 5 - 4 = 1.$
Let $\displaystyle u(s) = s^2+4^2$ then $\displaystyle \dfrac{du}{ds} = 2s~\implies~du = 2s \cdot ds$.
The integral becomes:
$\displaystyle \int \left(\dfrac s{\sqrt{s^2+4^2}} \right)ds = \dfrac12 \int \left(\dfrac {2s}{\sqrt{s^2+4^2}} \right)ds = \dfrac12 \int u^{-\frac12} du$
Can you take it from here?