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

Concept: To find this integral, we map it onto a standard, well-known trigonometric integration formula: \[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \] When the variable term \(x^2\) has a numerical leading coefficient multiplier other than $1$, we should first factor it completely out of the square root radical to make integration clean and straightforward.

Step 1: Manipulate the denominator term expression
The given integration expression is: \[ I = \int \frac{dx}{\sqrt{25 - 16x^2}} \] Let us factor out $16$ from the two terms inside the square root denominator: \[ 25 - 16x^2 = 16 \left( \frac{25}{16} - x^2 \right) \] Taking the square root of $16$ outside the radical gives: \[ \sqrt{25 - 16x^2} = \sqrt{16} \cdot \sqrt{\frac{25}{16} - x^2} = 4 \sqrt{\left(\frac{5}{4}\right)^2 - x^2} \]

Step 2: Rewrite the integral equation structure
Substitute this modified format back into our full integration problem: \[ I = \int \frac{dx}{4 \sqrt{\left(\frac{5}{4}\right)^2 - x^2}} \] Move the constant fraction coefficient outside the integral operator: \[ I = \frac{1}{4} \int \frac{dx}{\sqrt{\left(\frac{5}{4}\right)^2 - x^2}} \]

Step 3: Match coefficients with the standard formula
Comparing this expression directly to our standard integration template \(\int \frac{dx}{\sqrt{a^2 - x^2}}\), we can see that: \[ a = \frac{5}{4} \] Applying the inverse sine integration identity rule: \[ I = \frac{1}{4} \cdot \sin^{-1}\left(\frac{x}{\frac{5}{4}}\right) + C \] Simplifying the interior fraction expression: \[ I = \frac{1}{4} \sin^{-1}\left(\frac{4x}{5}\right) + C \] This precisely matches choice option (C).