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: Understanding the Concept:
This integral matches the standard inverse trigonometric form. To apply the formula, we must first ensure the coefficient of $x^2$ is 1 or express the denominator as a perfect square of a linear term.
Step 2: Key Formula or Approach:
1. Standard Formula: $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c$.
2. Generalized Formula: $\int \frac{dx}{\sqrt{a^2 - (mx)^2}} = \frac{1}{m} \sin^{-1}\left(\frac{mx}{a}\right) + c$.
Step 3: Detailed Explanation:
Rewrite the denominator: \[ \int \frac{dx}{\sqrt{4^2 - (5x)^2}} \] Here, $a = 4$ and the variable part is $5x$. Using the linear transformation rule (dividing by the coefficient of $x$): \[ = \frac{1}{5} \sin^{-1}\left(\frac{5x}{4}\right) + c \]
Step 4: Final Answer:
The integral is \( \frac{1}{5} \sin^{-1}\left(\frac{5x}{4}\right) + c \).