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Question

sin2xsin4x+4sin2x2dx

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Solution

sin2xsin4x+4sin2x2dx
Let sin2x=t
2sinxcosx=dt/dx
sin2x=dt/dx
sin2xdx=dt
dt(t)2+4t2 [(sin4x)=(sin2x)2]
Now completing whole square in denominator
dtt2+4t2+44
dtt24t+442
dtt22t2t+46=dtt(t2)2(t2)6
=dt(t2)26
=dt(t2)2(6)2 [1x2a2dx=logx2a2+x+c]
=log(t2)2(6)2+(t2)+c
t=sin2x
=log(sin2x2)2(6)2+(sin2x2)+c
=logsin4x4sin2x2+(sin2x2)+c

1175992_1291820_ans_cbfcccfd06f7429b981aac42af58b278.jpg

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