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If (sin2xcos2x)dx=12sin(2xa)+b then
A. a=5π4,bR
B.a=5π4,bR
C. a=π4,bR
D.None of these

Answer
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Hint: We consider the left-hand side of the expression and try to solve it by integrating step by step. In order to achieve the final expression, we need to do some adjustment operations in the expression. Once the final form of the expression is obtained, we can easily determine the required values of a and b.

Formula Used:
In order to solve this expression some important formulas are used that is,
sinaxdx=cosaxa+c
cosaxdx=sinaxa+c
Where, c is the constant of integration.
(sinAcosB+cosAsinB)=sin(A+B)


Complete step by step solution
Given:
The expression (sin2xcos2x)dx=12sin(2xa)+b.
Now, we will be solving the left hand side of the above expression, we have,(sin2xcos2x)dx=(cos2x2)(sin2x2)+c=12(cos2x+sin2x)+c
On rearranging the term by multiplying and dividing it with 2 we get,(sin2xcos2x)dx=12[cos2x(12)+sin2x(12)]+c=12(sin2xcosa+cos2xsina)+c
Where, we know that cosa=12 and sina=12.
Now, we know that the term (sin2xcosa+cos2xsina) can be written as the formula,(sin2xcosa+cos2xsina)=sin(2x+a)
We know that the values for both cos and sin are negative in the third quadrant so, we get,cosa=sina=12
So, the value of a will be equal to,
a=5π4
Then finally, the integral reduces to,
(sin2xcos2x)dx=12sin(2x+5π4)+c
Also, the above expression can also be written as:
(sin2xcos2x)dx=12sin(2xa)+b
Where, a and b are arbitrary constants, whose values are a=5π4 and bR (b belongs to real number).
The correct option is (B) that is a=5π4,bR.


Note: We have to reduce the expression in simplest form as given in the question that is why we had to do some rearrangements in the expression by multiplying-dividing with a number and using trigonometric relations.