I learnt something new today!

Thanks for improving my trigono skills. :-} Substitute $x=4\sin \theta$ such that $\, dx=4\cos \theta \, d\theta$ . $\large \mathfrak{K}=\displaystyle \int_0^{\frac{\pi}{2}} \dfrac{64\sin^3 \theta}{\sqrt{16-16\sin^2 \theta}} (4\cos \theta \, d\theta)$ $\large =64 \displaystyle \int_0^{\frac{\pi}{2}} \sin^3 \theta \, d\theta~~\color{#D61F06}{\small{(\because 1-\sin^2 \theta =\cos^2 \theta)}}$ Now using $\color{#D61F06}{\sin 3A=3\sin A-4\sin^3 A}$ $\color{#D61F06}{\implies \sin^3 A=\dfrac{3\sin A-\sin 3A}{4}}$ . $\large \mathfrak{K}=16\int_0^{\small{\frac{\pi}{2}}} (3\sin \theta-\sin 3\theta)\, d\theta$ $\large =16(-3\cos \theta +\dfrac{\cos 3\theta}{3})|_0^{\frac{\pi}{2}}$ $\huge =\dfrac{128}{3}$ $\huge \therefore ~128+3=\huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{131}}}}}$