Question

If $\int \frac{1}{\sqrt{9-16x^2}}dx=\alpha {\sin }^{-1}\left(\beta x\right)+c$, then $\alpha +\frac{1}{\beta }=$

• A. $1$
• B. $\frac{7}{12}$
• C. $\frac{19}{12}$
• D. $\frac{9}{12}$

Answer 1

The correct option is A $1$

$\int \frac{1}{\sqrt{9-16x^2}}dx=\alpha {\sin }^{-1}\left(\beta x\right)+c$ ...... $\left(i\right)$

Now, $\int \frac{1}{\sqrt{3^2-(4x{)}^2}}dx=\frac{1}{4}{\sin }^{-1}\left(\frac{4x}{3}\right)+c$

Comparing with $\left(i\right)$, we get
$\alpha =\frac{1}{4}$ and $\beta =\frac{4}{3}$

Then, $\alpha +\frac{1}{\beta }=\frac{1}{4}+\frac{3}{4}=1$

Hence, option A is correct.