∫(sin(tan^-1x))/(1+x^2) dx=

Let t = tan^-1 x ......(i)

$dt=\frac{dx}{1+x^2}$

it becomes ⇒ $∫sin\, t\, dt$

= - cot t + c ......(ii)

Substituting the value of t in equation (ii)

= -cos(tan^-1 x) + C

Option 1 is correct.