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Question

The value of ln3ln2xsinx2sinx2+sin(ln6x2)dx is

A
14In32
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B
12In32
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C
In32
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D
16In32
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Solution

The correct option is A 14In32
I= ln3ln2xsinx2sinx2+sin(ln6x2)dx

Put x2=t
=> 2xdx=dt
The integral will transform into
I= ln3ln21/2sintsint+sin(ln6t)dt
=> I= ln3ln21/2sin(ln6t)sint+sin(ln6t)dt

Adding both the equations, we get
2I= 12 ln3ln2sint+sin(ln6t)sint+sin(ln6t)dt

=> I= 14 ln3ln21dt

=> I= 14 [ln3ln2]

=> I= 14ln 32

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