Answer:
Option A
Explanation:
Let l= $\int\frac{1}{\sqrt{9-16x^{2}}}dx$
=$\int\frac{1}{\sqrt{(3)^2-(4x)^{2}}}dx=\frac{1}{4}\frac{1}{\sqrt{(\frac{3}{4})^{2}-(x)^{2}}}dx$
= $\frac{1}{4}\sin^{-1}\frac{4x}{3}+C$
$\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} dx=\sin^{-1}\frac{x}{a}+C\right]$
But , it is given that
$\int\frac{1}{\sqrt{9-16x^{2}}}dx=\alpha \sin ^{-1}(\beta x)+C$
$\therefore$ $\alpha \sin ^{-1}(\beta x)+C=\frac{1}{4}\sin^{-1}(\frac{4}{3}x)+C$
On comparing both sides, we get
$\alpha =\frac{1}{4}$ and $\beta =\frac{4}{3}$
$\therefore$ $\alpha+\frac{1}{\beta} =\frac{1}{4}+\frac{1}{4}=\frac{1}{4}+\frac{3}{4}=1$
