Definite Integration Test

$\begin{aligned} I & = \int_0^{\frac {\sqrt 5}2} \frac {60}{(1+4x^2)\sqrt{25-20x^2}}\ dx & \small \color{#3D99F6} \text{Let }\sin \theta = \frac {2x}{\sqrt 5} \implies \cos \theta \ d \theta = \frac 2{\sqrt 5} \ dx \\ & = \int_0^\frac \pi 2 \frac {6\sqrt 5 \cos \theta}{(1+5\sin^2 \theta)\sqrt{1-\sin^2 \theta}}\ d\theta \\ & = \int_0^\frac \pi 2 \frac {6\sqrt 5}{1+5\sin^2 \theta}\ d\theta & \small \color{#3D99F6} \text{Since }\cos 2\theta = 1-2\sin^2 \theta \\ & = \int_0^\frac \pi 2 \frac {12\sqrt 5}{7-5\cos 2\theta}\ d\theta & \small \color{#3D99F6} \text{Let }t = \tan \theta \implies dt = \sec^2 \theta \ d \theta \\ & = \int_0^\infty \frac {12\sqrt 5}{7-5 \left(\frac {1-t^2}{1+t^2}\right)}\times \frac {dt}{1+t^2} \\ & = \int_0^\infty \frac {6\sqrt 5}{1+6t^2} \ dt & \small \color{#3D99F6} \text{Let }\sqrt 6 t = \tan \phi \implies \sqrt 6 dt = \sec^2 \phi \ d \phi \\ & = \sqrt {30} \int_0^\frac \pi 2 \frac {\sec^2 \phi}{1+\tan^2 \phi} \ d \phi \\ & = \sqrt {30} \int_0^\frac \pi 2 \ d \phi \\ & = \sqrt {30} \phi \ \bigg|_0^\frac \pi 2 = \pi \sqrt{\frac {15}2} \end{aligned}$

Therefore, $ab = 15\times 2 = \boxed {30}$ .

$\int_0^{\frac{\sqrt{5}}{2}} \frac{60}{\left(4 x^2+1\right) \sqrt{25-20 x^2}} \, dx \Rightarrow \sqrt{30} \tan ^{-1}\left(\frac{2 \sqrt{6} x}{\sqrt{5-4 x^2}}\right)$ $\large|_{x=0}^{x=\frac{\sqrt{5}}{2}}$

$\sqrt{30} \tan ^{-1}\left(\frac{2 \sqrt{6} x}{\sqrt{5-4 x^2}}\right)|{x=0}\Rightarrow 0$

$\underset{x\to \left(\frac{\sqrt{5}}{2}\right)^-}{\text{lim}}\sqrt{30} \tan ^{-1}\left(\frac{2 \sqrt{6} x}{\sqrt{5-4 x^2}}\right)= \sqrt{\frac{15}{2}} \pi$

The limit has to be taken from below as the expression can not be evaluated from above in $\mathbb{R}$ and taking the limit from this direction is sufficient.

Taking the limit is necessary to the evaluation as the direct substitution is indeterminate.

$I = \frac{60}{\sqrt{5}} \int \limits_0^{\frac{\sqrt{5}}{2}} \! \frac{\mathrm{d}x}{1+4x^2 \sqrt{5-4x^2} }$ . Let $x = \frac{\sqrt{5}}{2}y$ ,

$= 6\sqrt{5} \int \limits_0^1 \! \frac{\mathrm{d}y}{1+5y^2 \sqrt{1-y^2} }$ . Let $y = \sin{\theta}$ ,

$= 6\sqrt{5} \int \limits_0^\frac{\pi}{2} \! \frac{\mathrm{d}\theta}{1+5\sin^2{\theta}} = \int \limits_0^\frac{\pi}{2} \! \frac{\mathrm{d}\theta}{\cos^2{\theta}+6\sin^2{\theta}} = \int \limits_0^\frac{\pi}{2} \! \frac{\sec^2{\theta} \, \mathrm{d}\theta}{1+6\tan^2{\theta}}$ . Let $\phi = \sqrt{6}\tan{\theta}$ .

$= \sqrt{30} \int \limits_0^\infty \! \frac{\mathrm{d}\phi}{1+\phi^2}$ . Let $\phi = \tan{\zeta}$ .

$= \sqrt{30} \int \limits_0^\frac{\pi}{2} \mathrm{d}\zeta = \sqrt{\frac{15}{2} } \pi$ .