Question:
\(\int \frac{1}{\sqrt{2x-x^2}}dx =\)
\(\int \frac{1}{\sqrt{2x-x^2}}dx =\)
Show Hint
Complete the square under the root to reach the sin inverse standard form.
Updated On: Oct 1, 2026
- \(sin^{-1}(x-1)+c\)
- \(cos^{-1}(x-1)+c\)
- \(tan^{-1}(x-1)+c\)
- \(sin^{-1}x+c\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Complete the square so that the root has the form \(\sqrt{a^2-u^2}\).
Step 2: Complete the square
\[ 2x-x^2=1-(x^2-2x+1)=1-(x-1)^2 \]
Step 3: Integrate
Let \(u=x-1\), \(du=dx\):
\[ \int\frac{du}{\sqrt{1-u^2}}=\sin^{-1}u+c=\sin^{-1}(x-1)+c \]
Step 4: Check the options
Differentiating \(\cos^{-1}(x-1)\) gives a negative sign, so (B) is not an antiderivative. \(\tan^{-1}(x-1)\) has a different form, and \(\sin^{-1}x\) has the wrong argument. The answer is (A).
Final Answer:
Completing the square gives sin inverse of (x - 1), option (A). \[ \boxed{\sin^{-1}(x-1)+c} \]
Complete the square so that the root has the form \(\sqrt{a^2-u^2}\).
Step 2: Complete the square
\[ 2x-x^2=1-(x^2-2x+1)=1-(x-1)^2 \]
Step 3: Integrate
Let \(u=x-1\), \(du=dx\):
\[ \int\frac{du}{\sqrt{1-u^2}}=\sin^{-1}u+c=\sin^{-1}(x-1)+c \]
Step 4: Check the options
Differentiating \(\cos^{-1}(x-1)\) gives a negative sign, so (B) is not an antiderivative. \(\tan^{-1}(x-1)\) has a different form, and \(\sin^{-1}x\) has the wrong argument. The answer is (A).
Final Answer:
Completing the square gives sin inverse of (x - 1), option (A). \[ \boxed{\sin^{-1}(x-1)+c} \]
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