integral o to π/2 tanx/1+m²tan²x....Solve this......Provide solution fast

0 to π/2 tanxdx/(1 +m^2tan^2x)=0 to π/2 (sinx/cosx)dx/((cos^2x+m^2 sin^2x)/cos^2x)=0 to π/2 sinxcosxdx/(cos^2x + m^2(1-cos^2x))Put cosx=t dt =-sinxdx and limit changes to 1 to 01 to 0 -tdt/(t^2 + m^2(1-t^2))= 0 to 1 tdt/(t^2(1-m^2)+m^2)= 1/2(1-m^2) 0 to 1 2t(1-m^2)dt/(t^2(1-m^2)+m^2)=1/2(1-m^2)[ln(t^2(1-m^2)+m^2)] from 0 to 1=1/2(1-m^2)[0-2lnm]=lnm/(m^2-1)