Question

\[\int\limits_0^a \frac{1}{x + \sqrt{a^2 - x^2}} dx\]

Answer 1

We have, 

\[ I = \int_0^a \frac{1}{x + \sqrt{a^2 - x^2}} d x\]

Putting \[x = a \sin \theta\]

\[ \Rightarrow dx = a \cos \theta d\theta\]

\[\text{When }x \to 0; \theta \to 0 \]

\[\text{And }x \to a; \theta \to \frac{\pi}{2}\]

\[ \therefore I = \int_0^\frac{\pi}{2} \frac{a \cos \theta}{a \sin \theta + \sqrt{a^2 - \left( a \sin \theta \right)^2}}d\theta\]

\[ = \int_0^\frac{\pi}{2} \frac{a \cos \theta}{a \sin \theta + a \cos \theta}d\theta\]

\[I = \int_0^\frac{\pi}{2} \frac{\cos \theta}{\sin \theta + \cos \theta}d\theta . . . . . \left( 1 \right)\]

\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\cos \left( \frac{\pi}{2} - \theta \right)}{\sin \left( \frac{\pi}{2} - \theta \right) + \cos \left( \frac{\pi}{2} - \theta \right)}d\theta\]

\[ = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\cos \theta + \sin \theta}d\theta\]

\[I = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\sin \theta + \cos \theta}d\theta . . . . . \left( 2 \right)\]

\[\text{By adding (1) and (2), we get}\]

\[2I = \int_0^\frac{\pi}{2} \frac{\cos \theta + \sin \theta}{\sin \theta + \cos \theta}d\theta \]

\[ \Rightarrow 2I = \int_0^\frac{\pi}{2} d\theta \]

\[ \Rightarrow 2I = \left[ \theta \right]_0^\frac{\pi}{2} \]

\[ \Rightarrow 2I = \frac{\pi}{2}\]

\[ \Rightarrow I = \frac{\pi}{4}\]