\(\int\displaystyle\frac{\sin(\tan^{-1}x)}{1+x^2}\,dx=\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\cos(\tan^{-1}x)+C\)
The factor \(\frac{1}{1+x^2}\) is the derivative of \(\tan^{-1}x\), so substitute \(t = \tan^{-1}x\):
\( \displaystyle dt = \frac{dx}{1+x^2} \)
\( \displaystyle \int\frac{\sin(\tan^{-1}x)}{1+x^2}\,dx = \int\sin t\,dt \)
\(= -\cos t + C\)
\(= -\cos(\tan^{-1}x) + C\)
NCERT (2026-27 edition):
Integration by Substitution
