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Question

The value of dx(3sinx4cosx)2 is
(where C is integration constant)

A
131(3tanx4)+C
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B
431(3cotx4)+C
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C
131(3cotx4)+C
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D
431(3tanx4)+C
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Solution

The correct option is A 131(3tanx4)+C
I=dx(3sinx4cosx)2
Dividing numerator and denominator by cos2x
=sec2xdx(3tanx4)2

Put 3tanx4=t
3sec2xdx=dt
I=dt3t2=131(3tanx4)+C

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