∫sin2xsin4x+4sin2x−2dx
Let sin2x=t
2sinxcosx=dt/dx
sin2x=dt/dx
sin2xdx=dt
∫dt√(t)2+4t−2 [(sin4x)=(sin2x)2]
Now completing whole square in denominator
∫dt√t2+4t−2+4−4
∫dtt2−4t+4−4−2
∫dt√t2−2t−2t+4−6=∫dt√t(t−2)−2(t−2)−6
=∫dt√(t−2)2−6
=∫dt√(t−2)2−(√6)2 [1√x2−a2dx=log∣∣√x2−a2+x∣∣+c]
=log∣∣∣√(t−2)2−(√6)2+(t−2)∣∣∣+c
t=sin2x
=log∣∣∣√(sin2x−2)2−(√6)2+(sin2x−2)∣∣∣+c
=log∣∣√sin4x−4sin2x−2+(sin2x−2)∣∣+c