Question

If \[ \int \frac{dx}{\sqrt{16 - 9x^2}} = A \sin^{-1}(Bx) + C, { where } C { is an arbitrary constant, then } A + B = \]

• A. 4
• B. 0
• C. \(\frac{3}{4}\)
• D. 1
• E. \(\frac{1}{4}\)

Hint:

Use the standard integral forms to simplify the problem and solve for constants.

Answer 1

Answer: (E)

The given equation is: \[ \int \frac{dx}{\sqrt{16 - 9x^2}} = A \sin^{-1}(Bx) + C \] To solve this, we recognize that this is a standard integral of the form: \[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) \] For our case, \(a^2 = 16\) and \(x^2\) is replaced by \(9x^2\), so we use substitution. 
After solving, we find that \(A = \frac{1}{4}\) and \(B = 3\). 
Thus, \(A + B = \frac{1}{4} + 3 = \frac{1}{4}\). Therefore, the correct answer is \(\boxed{\frac{1}{4}}\).