Question:
\[
\int \frac{dx}{\sqrt{16-25x^2}}=
\]
\[
\int \frac{dx}{\sqrt{16-25x^2}}=
\]
Show Hint
Convert \(\sqrt{16-25x^2}\) into \(\sqrt{4^2-(5x)^2}\), then use the inverse sine formula.
Updated On: May 7, 2026
- \(\frac{1}{5}\sin^{-1}\left(\frac{5x}{4}\right)+c\)
- \(\sin^{-1}\left(\frac{5x}{4}\right)+c\)
- \(\frac{1}{5}\sin^{-1}\left(\frac{x}{4}\right)+c\)
- \(\frac{1}{5}\sin^{-1}\left(\frac{4x}{5}\right)+c\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Use the standard formula:
\[
\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+c
\]
Step 1: Given: \[ \int \frac{dx}{\sqrt{16-25x^2}} \]
Step 2: Rewrite the expression inside the square root. \[ 16-25x^2=4^2-(5x)^2 \] So: \[ \sqrt{16-25x^2} = \sqrt{4^2-(5x)^2} \]
Step 3: Put: \[ u=5x \] Then: \[ du=5dx \] \[ dx=\frac{du}{5} \]
Step 4: Substitute in the integral. \[ \int \frac{dx}{\sqrt{16-25x^2}} = \frac{1}{5}\int \frac{du}{\sqrt{16-u^2}} \]
Step 5: Apply the standard formula. \[ \frac{1}{5}\int \frac{du}{\sqrt{4^2-u^2}} = \frac{1}{5}\sin^{-1}\left(\frac{u}{4}\right)+c \]
Step 6: Substitute \(u=5x\). \[ = \frac{1}{5}\sin^{-1}\left(\frac{5x}{4}\right)+c \] Therefore, \[ \boxed{\frac{1}{5}\sin^{-1}\left(\frac{5x}{4}\right)+c} \]
Step 1: Given: \[ \int \frac{dx}{\sqrt{16-25x^2}} \]
Step 2: Rewrite the expression inside the square root. \[ 16-25x^2=4^2-(5x)^2 \] So: \[ \sqrt{16-25x^2} = \sqrt{4^2-(5x)^2} \]
Step 3: Put: \[ u=5x \] Then: \[ du=5dx \] \[ dx=\frac{du}{5} \]
Step 4: Substitute in the integral. \[ \int \frac{dx}{\sqrt{16-25x^2}} = \frac{1}{5}\int \frac{du}{\sqrt{16-u^2}} \]
Step 5: Apply the standard formula. \[ \frac{1}{5}\int \frac{du}{\sqrt{4^2-u^2}} = \frac{1}{5}\sin^{-1}\left(\frac{u}{4}\right)+c \]
Step 6: Substitute \(u=5x\). \[ = \frac{1}{5}\sin^{-1}\left(\frac{5x}{4}\right)+c \] Therefore, \[ \boxed{\frac{1}{5}\sin^{-1}\left(\frac{5x}{4}\right)+c} \]
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