Question

$\underset{0}{\overset{a}{\int }}\frac{1}{x+\sqrt{a^2-x^2}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^a\frac{1}{x+\sqrt{a^2-x^2}}dx \\ \mathrm{Putting}x=a\sin \theta \\ \Rightarrow dx=a\cos \theta d\theta \\ \mathrm{When}x\to 0;\theta \to 0 \\ \mathrm{And}x\to a;\theta \to \frac{\mathrm{\pi }}{2} \\ \therefore I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{a\cos \theta }{a\sin \theta +\sqrt{a^2-{\left(a\sin \theta \right)}^2}}d\theta \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{a\cos \theta }{a\sin \theta +a\cos \theta }d\theta \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \theta }{\sin \theta +\cos \theta }d\theta .....\left(1\right) \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\theta \right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\theta \right)+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\theta \right)}d\theta \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \theta }{\cos \theta +\sin \theta }d\theta \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \theta }{\sin \theta +\cos \theta }d\theta .....\left(2\right) \\ \mathrm{By}\mathrm{adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \theta \mathit{}+\sin \theta }{\sin \theta +\cos \mathit{}\theta }d\theta \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}d\theta \\ \Rightarrow 2I={\left[\theta \right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{2} \\ \Rightarrow I=\frac{\mathrm{\pi }}{4} \end{gathered}$$