Question

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

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

Answer 1

Answer: (A)

1

Answer 2

The given integral is $\int \frac{1}{\sqrt{9-16x^2}}dx$. We need to transform the integrand into a standard form, specifically $\frac{1}{\sqrt{a^2-x^2}}$.

Step 1: Simplify the denominator

Factor out 16 from the expression inside the square root: $\sqrt{9-16x^2} = \sqrt{16\left(\frac{9}{16}-x^2\right)}$ $= \sqrt{16} \sqrt{\left(\frac{3}{4}\right)^2-x^2}$ $= 4\sqrt{\left(\frac{3}{4}\right)^2-x^2}$

Now, substitute this back into the integral: $\int \frac{1}{4\sqrt{\left(\frac{3}{4}\right)^2-x^2}}dx = \frac{1}{4} \int \frac{1}{\sqrt{\left(\frac{3}{4}\right)^2-x^2}}dx$

Step 2: Apply the standard integration formula

The standard integral formula for $\int \frac{1}{\sqrt{a^2-x^2}}dx = \sin^{-1}\left(\frac{x}{a}\right) + C$. In our case, $a = \frac{3}{4}$. So, the integral becomes: $\frac{1}{4} \sin^{-1}\left(\frac{x}{\frac{3}{4}}\right) + c$ $= \frac{1}{4} \sin^{-1}\left(\frac{4x}{3}\right) + c$

Step 3: Compare with the given form

The problem states that $\int \frac{1}{\sqrt{9-16x^2}}dx = \alpha\sin^{-1}(\beta x)+c$. Comparing our result $\frac{1}{4} \sin^{-1}\left(\frac{4x}{3}\right) + c$ with the given form, we can identify the values of $\alpha$ and $\beta$: $\alpha = \frac{1}{4}$ $\beta = \frac{4}{3}$

Step 4: Calculate $\alpha+\frac{1}{\beta}$

First, find $\frac{1}{\beta}$: $\frac{1}{\beta} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$

Now, calculate $\alpha+\frac{1}{\beta}$: $\alpha+\frac{1}{\beta} = \frac{1}{4} + \frac{3}{4} = \frac{1+3}{4} = \frac{4}{4} = 1$

The final answer is 1.