Question

$\underset{0}{\overset{a}{\int }}\frac{x}{\sqrt{a^2+x^2}}dx=$ _______________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{a}{\int }}\frac{x}{\sqrt{a^2+x^2}}dx \\ \mathrm{Let}, \\ {a}^2+x^2=t \\ \Rightarrow d\left(a^2+x^2\right)=dt \\ \Rightarrow 2xdx=dt \\ \mathrm{Also},\mathrm{if}x=0,t=a^2 \\ \mathrm{if}x=a,t=2a^2 \\ \mathrm{Thus}, \\ I=\frac{1}{2}\underset{a^2}{\overset{2a^2}{\int }}\frac{1}{\sqrt{t}}dt \\ =\frac{1}{2}{\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]}_{a^2}^{2a^2} \\ =\frac{1}{2}{\left[\frac{t^{{\displaystyle \frac{1}{2}}}}{{\displaystyle \frac{1}{2}}}\right]}_{a^2}^{2a^2} \\ =\frac{1}{2}{\left[2\sqrt{t}\right]}_{a^2}^{2a^2} \\ =\frac{1}{2}\left[2\sqrt{2a^2}-2\sqrt{a^2}\right] \\ =\frac{1}{2}\left[2a\sqrt{2}-2a\right] \\ =\left[a\sqrt{2}-a\right] \\ =a\left(\sqrt{2}-1\right) \end{gathered}$$

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Hence, $\underset{0}{\overset{a}{\int }}\frac{x}{\sqrt{a^2+x^2}}dx=\underline{a\left(\sqrt{2}-1\right)}.$