Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}x=a\tan t.\mathrm{Then},dx=a{\sec }^{2}tdt \\ \mathrm{When}x=0,t=0\mathrm{and}x=a,t=\frac{\mathrm{\pi }}{4} \\ \therefore I={\int }_0^a\frac{x}{\sqrt{a^2+x^2}}\mathit{d}x \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{a\tan t}{\sqrt{a^2+a^2{\tan }^{2}t}}a{\sec }^{2}t\mathit{d}t \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\left(a\tan t\right)a{\sec }^{2}t}{a\sec t}dt \\ ={\int }_0^{\frac{\mathrm{\pi }}{4}}a\tan t\sec tdt \\ =a{\left[\sec t\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ =a\left(\sqrt{2}-1\right) \end{gathered}$$