Question:
If \(\int \frac{dx}{\sqrt{2ax-x^2}} = f(g(x))+c\), where \(c\) is constant of integration, then \(f(x),g(x)\) are respectively equal to
If \(\int \frac{dx}{\sqrt{2ax-x^2}} = f(g(x))+c\), where \(c\) is constant of integration, then \(f(x),g(x)\) are respectively equal to
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Complete the square: $2ax-x^2=a^2-(x-a)^2$.
Updated On: Oct 1, 2026
- \(sin^{-1}x,\frac{x-a}{a}\)
- \(sin^{-1}x,\frac{x+a}{a}\)
- \(tan^{-1}x,\frac{x-a}{a}\)
- \(tan^{-1}x,\frac{x+a}{a}\)
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The Correct Option is A
Solution and Explanation
Step 1: Complete the square
\(2ax-x^2=a^2-(x-a)^2\).
Step 2: Use the standard form
\(\int\frac{dx}{\sqrt{a^2-(x-a)^2}}=\sin^{-1}\frac{x-a}{a}+c\).
Step 3: Read \(f\) and \(g\)
So \(f(x)=\sin^{-1}x\) and \(g(x)=\frac{x-a}{a}\). Option (A).
Final Answer:
\(f=\sin^{-1}x\) and \(g=\frac{x-a}{a}\), option (A). \[ \boxed{\text{(A)}} \]
\(2ax-x^2=a^2-(x-a)^2\).
Step 2: Use the standard form
\(\int\frac{dx}{\sqrt{a^2-(x-a)^2}}=\sin^{-1}\frac{x-a}{a}+c\).
Step 3: Read \(f\) and \(g\)
So \(f(x)=\sin^{-1}x\) and \(g(x)=\frac{x-a}{a}\). Option (A).
Final Answer:
\(f=\sin^{-1}x\) and \(g=\frac{x-a}{a}\), option (A). \[ \boxed{\text{(A)}} \]
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