Question

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

Answer 1

\[Let\, I = \int\limits_0^a x \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} dx\]

\[Consider\, x^2 = a^2 \cos2\theta\]

\[ \Rightarrow 2x\ dx = - 2 a^2 \sin2\theta d\theta\]

\[ \Rightarrow x\ dx = - a^2 \sin2\theta d\theta\]

\[When\, x \to 0 ; \theta \to \frac{\pi}{4} and\ x \to a ; \theta \to 0\]

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

\[I = \int_\frac{\pi}{4}^0 - a^2 \sin2\theta \sqrt{\frac{a^2 - a^2 \cos2\theta}{a^2 + a^2 \cos2\theta}} d\theta\]

\[ = \int_\frac{\pi}{4}^0 - a^2 \sin2\theta \tan\theta d\theta\]

\[ = a^2 \int_0^\frac{\pi}{4} 2 \sin\theta \cos\theta \frac{\sin\theta}{\cos\theta} d\theta\]

\[ = a^2 \int_0^\frac{\pi}{4} 2 \sin^2 \theta d\theta\]

\[ = a^2 \int_0^\frac{\pi}{4} \left[ 1 - \cos 2\theta \right] d\theta\]

\[ = a^2 \left[ \theta - \frac{\sin2\theta}{2} \right]_0^\frac{\pi}{4} \]

\[ = a^2 \left[ \frac{\pi}{4} - \frac{1}{2} \right]\]