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(sin 2xcos 2x)dx=12sin (2xa)+b, then 

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a
a=5π4,b∈R
b
a=−5π4,b∈R
c
a=π4,b∈R
d
None of these

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detailed solution

Correct option is B

∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x−a)+b⇒ 2∫12sin⁡2x−12cos⁡2xdx=12sin⁡(2x−a)+b⇒−2∫12cos⁡2x−12sin⁡2xdx=12sin⁡(2x−a)+b⇒ −2∫cos⁡2x+π4dx=12sin⁡(2x−a)+b⇒  −22sin⁡2x+π4+C=12sin⁡(2x−a)+b⇒ 12sin⁡2x+5π4+C=12sin⁡(2x−a)+b⇒ a=−5π4,b∈R


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