Question:
$\displaystyle \int \frac{\sin(\cot^{-1}x)}{1+x^2} \, dx$ 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
- $-\cos(\cot^{-1}x) + C$
- $\cos(\cot^{-1}x) + C$
- $\frac{\cos(\cot^{-1}x)}{1+x^2} + C$
- $\frac{\cos(\cot^{-1}x)}{2} + C$
- $-\frac{\cos(\cot^{-1}x)}{1+x^2} + C$
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The Correct Option is B
Solution and Explanation
Concept:
• Use substitution: $t = \cot^{-1}x$
Step 1: Substitute
\[ t = \cot^{-1}x \Rightarrow \frac{dt}{dx} = -\frac{1}{1+x^2} \] \[ dx = -(1+x^2)dt \]
Step 2: Substitute in integral
\[ \int \frac{\sin(\cot^{-1}x)}{1+x^2} dx = \int \sin t \cdot (-dt) \] \[ = -\int \sin t \, dt \]
Step 3: Integrate
\[ = \cos t + C \]
Step 4: Back substitute
\[ = \cos(\cot^{-1}x) + C \] Final Conclusion:
Option (B)
• Use substitution: $t = \cot^{-1}x$
Step 1: Substitute
\[ t = \cot^{-1}x \Rightarrow \frac{dt}{dx} = -\frac{1}{1+x^2} \] \[ dx = -(1+x^2)dt \]
Step 2: Substitute in integral
\[ \int \frac{\sin(\cot^{-1}x)}{1+x^2} dx = \int \sin t \cdot (-dt) \] \[ = -\int \sin t \, dt \]
Step 3: Integrate
\[ = \cos t + C \]
Step 4: Back substitute
\[ = \cos(\cot^{-1}x) + C \] Final Conclusion:
Option (B)
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