Question

$\int \frac{dx}{\sqrt{16-25x^{2}}} =$

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

Hint:

For $\int \frac{dx}{\sqrt{a^2 - (bx)^2}}$, the result is always $\frac{1}{b}\sin^{-1}(\frac{bx}{a}) + c$.

Answer 1

The Correct Option is A

Step 1: Concept
This matches the standard integral form $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}(\frac{u}{a}) + c$.

Step 2: Meaning
Factor out 25 or rewrite as $\int \frac{1}{\sqrt{4^2 - (5x)^2}} dx$.

Step 3: Analysis
Let $u = 5x$, then $du = 5 dx \implies dx = \frac{1}{5} du$. The integral is $\frac{1}{5} \int \frac{du}{\sqrt{4^2 - u^2}}$.

Step 4: Conclusion
Integrating gives $\frac{1}{5} \sin^{-1}(\frac{u}{4}) + c = \frac{1}{5} \sin^{-1}(\frac{5x}{4}) + c$.
Final Answer: (A)