Question:medium

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

Show Hint

Convert \(\sqrt{16-25x^2}\) into \(\sqrt{4^2-(5x)^2}\), then use the inverse sine formula.
  • \(\frac{1}{5}\sin^{-1}\left(\frac{5x}{4}\right)+c\)
  • \(\sin^{-1}\left(\frac{5x}{4}\right)+c\)
  • \(\frac{1}{5}\sin^{-1}\left(\frac{x}{4}\right)+c\)
  • \(\frac{1}{5}\sin^{-1}\left(\frac{4x}{5}\right)+c\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The integrand is of the form \( \frac{1}{\sqrt{a^2 - (kx)^2}} \).
This structure is typical of the derivative of an inverse sine function.
To solve it, we can either use a direct formula for standard integrals or use a substitution.
Step 2: Key Formula or Approach:
1. Standard Integral: \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(x/a) + C \).
2. If \( x \) is replaced by \( kx \), then \( \int \frac{dx}{\sqrt{a^2 - (kx)^2}} = \frac{1}{k} \sin^{-1}(kx/a) + C \).
Step 3: Detailed Explanation:

Rewrite the expression inside the root as a difference of squares:
\( 16 = 4^2 \) and \( 25x^2 = (5x)^2 \).
The integral is \( \int \frac{dx}{\sqrt{4^2 - (5x)^2}} \).

Use substitution to make it look like the standard form:
Let \( u = 5x \). Then \( du = 5 dx \implies dx = du/5 \).

Substitute into the integral:
\[ \int \frac{du/5}{\sqrt{4^2 - u^2}} = \frac{1}{5} \int \frac{du}{\sqrt{4^2 - u^2}} \]

Apply the standard formula with \( a = 4 \):
\[ \frac{1}{5} \sin^{-1} \left(\frac{u}{4}\right) + c \]

Substitute back \( u = 5x \):
\[ \frac{1}{5} \sin^{-1} \left(\frac{5x}{4}\right) + c \]

Step 4: Final Answer:
The integral is \( \frac{1}{5} \sin^{-1} \left(\frac{5x}{4}\right) + c \).
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