Question

The indefinite integral \(\int \frac{dx}{\sqrt{25 - 16x^2}}\) is equal to:

• A. \(\frac{1}{5} \sin^{-1}(4x) + C\)
• B. \(\frac{1}{25} \sin^{-1}(16x) + C\)
• C. \(\frac{1}{4} \sin^{-1}\left(\frac{4x}{5}\right) + C\)
• D. \(\frac{1}{16} \sin^{-1}\left(\frac{4x}{5}\right) + C\)

Hint:

Alternatively, you can apply a simple linear substitution rule: if \(\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a})\), then \(\int \frac{dx}{\sqrt{a^2 - (kx)^2}} = \frac{1}{k}\sin^{-1}(\frac{kx}{a})\). Here, \(a=5\) and \(k=4\), giving \(\frac{1}{4}\sin^{-1}(\frac{4x}{5})\) instantly.

Answer 1

The Correct Option is C