Question

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

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

Hint:

Alternatively, you can factor out 16 from the square root: \(\sqrt{16(\frac{25}{16} - x^2)} = 4\sqrt{(\frac{5}{4})^2 - x^2}\).
Remember that the derivative of \(\sin^{-1}(kx)\) includes a factor of \(k\) due to the chain rule, which is why we need \(1/k\) in the integral.
Always include the constant of integration \(C\) for indefinite integrals.

Answer 1

The Correct Option is C

Concept:
• Use the standard integral formula: \(\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C\).
• If there is a coefficient with \(x^2\), use substitution or factor out the coefficient.

Step 1: Rewrite the integrand in standard form
The integral is \(I = \int \frac{dx}{\sqrt{25 - 16x^2}}\).
We can express the terms as squares:
\[ I = \int \frac{dx}{\sqrt{5^2 - (4x)^2}} \]

Step 2: Apply substitution
Let \(4x = t\).
Then, differentiating both sides: \(4 dx = dt \implies dx = \frac{dt}{4}\).
Substitute these into the integral:
\[ I = \int \frac{\frac{dt}{4}}{\sqrt{5^2 - t^2}} = \frac{1}{4} \int \frac{dt}{\sqrt{5^2 - t^2}} \]

Step 3: Integrate using the standard formula
Using \(\int \frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\left(\frac{t}{a}\right) + C\) with \(a = 5\):
\[ I = \frac{1}{4} \left[ \sin^{-1}\left(\frac{t}{5}\right) \right] + C \] Substitute back \(t = 4x\):
\[ I = \frac{1}{4} \sin^{-1}\left(\frac{4x}{5}\right) + C \] This matches option (C).