Question:medium
$\displaystyle \int \frac{\sin(\cot^{-1}x)}{1+x^2} \, dx$ is equal to:
$\displaystyle \int \frac{\sin(\cot^{-1}x)}{1+x^2} \, dx$ is equal to:
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Inverse trig substitution simplifies complex integrals.
Updated On: Apr 24, 2026
- $-\cos(\cot^{-1}x) + C$
- $\cos(\cot^{-1}x) + C$
- $\frac{\cos(\cot^{-1}x)}{1+x^2} + C$
- $\frac{\cos(\cot^{-1}x)}{2} + C$
- $-\frac{\cos(\cot^{-1}x)}{1+x^2} + C$
Show Solution
The Correct Option is B
Solution and Explanation
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