Question

\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

Answer 1

\[Let\, I = \int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

\[Consider\, x = a \cos 2y\ Then\ y = \frac{1}{2} \cos^{- 1} \left( \frac{x}{a} \right)\]

\[ \Rightarrow dx = - 2a \sin 2y\ dy\]

\[When\, x \to - a ; y \to \frac{\pi}{2}\ and\ x\ \to a ; y \to 0\]

\[\text{Now, integral becomes}, \]

\[ I = \int_\frac{\pi}{2}^0 - 2a \sin 2y\sqrt{\frac{a - a \cos 2y}{a + a \cos 2y}} dy\]

\[ = \int_0^\frac{\pi}{2} 2a \sin 2y \tan\ y\ dy\]

\[ = 2a \int_0^\frac{\pi}{2} 2\sin y \cos y \frac{\sin y}{\cos y}\ dy\]

\[ = 2a \int_0^\frac{\pi}{2} 2 \sin^2\ y\ dy\]

\[ = 2a \int_0^\frac{\pi}{2} \left( 1 - \cos 2y \right) dy\]

\[ = 2a \left[ y - \frac{\sin 2y}{2} \right]_0^\frac{\pi}{2} \]

\[ = 2a \left[ \frac{\pi}{2} - \frac{\sin 2y}{2} \right]_0^\frac{\pi}{2} \]

\[ = \pi a\]