\(\int\displaystyle\frac{\sin\left(\tan^{-1}x\right)}{1+x^2}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\cos\left(\tan^{-1}x\right)+C\)
Since \(\frac{1}{1+x^2}\) is the derivative of \(\tan^{-1}x\), the substitution \(t = \tan^{-1}x\) removes the fraction.
\( \displaystyle t = \tan^{-1}x,\quad dt = \frac{dx}{1+x^2} \)
\( \displaystyle \int\frac{\sin\left(\tan^{-1}x\right)}{1+x^2}\,dx = \int\sin t\,dt \)
\( = -\cos t+C \)
\( = -\cos\left(\tan^{-1}x\right)+C \)