Question

Evaluate $\stackrel{3}{\underset{0}{\int }}\frac{x}{\sqrt{x^2+16}}dx$.

Answer 1

${\int }_0^3\frac{x}{\sqrt{x^2+16}}dx$

Let

$x^2+16=t\{\begin{array}{c}\text{when}x=0\Rightarrow t=16\\ \text{when}x=3\Rightarrow t=25\end{array}$

Differentiating above equation w.r.t. $x$, we have

$2xdx=dt$

$xdx=\frac{dt}{2}$

Now, the integration will be in the form-

${\int }_{16}^{25}\frac{dt}{2\sqrt{t}}$

$=\frac{1}{2}\int t^{-\frac{1}{2}}dt$

$=\frac{1}{2}{\left[\sqrt{t}\right]}_{16}^{25}$

$=\frac{1}{2}[\sqrt{25}-\sqrt{16}]$

$=\frac{1}{2}(5-4)=\frac{1}{2}$

Hence the value of ${\int }_0^3\frac{x}{\sqrt{x^2+16}}dx$ is $\frac{1}{2}$.