tan X/2 = t ,sinx = 2t/1+t² ; cos x = 1–t² / 1+t², x = 2 tan^–1t; dx = 2dt/1+t²

how we derive dx = 2dt/1+t²

thanks

sinx = 2t/1+t²

cosxdx = ((1+t^2)2 - 4t^2)/(1+t^2)^2 * dt

1-t^2/1+t^2 *dx = 2* (1-t^2)/(1+t^2)^2 * dt

cancelling on Lhs and Rhs

dx = 2dt/1+t^2

If you like my answer please approve by clicking on the "Yes" given below!!!