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The value of I=2sinxsin(xπ/4)dx is

 

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a
x+log⁡|cos⁡(x−π/4)|+C
b
x−log⁡|sin⁡(x−π/4)|+C
c
x+log⁡|sin⁡(x−π/4)|+C
d
x−log⁡|cos⁡(x−π/4)|+C

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detailed solution

Correct option is C

Put x−π/4=tI=2∫sin⁡(π/4+t)sin⁡tdt=∫(cot⁡t+1)dt=t+log⁡|sin⁡t|+C′=x+log⁡|sin⁡(x−π/4)|+C,C=C′+π/4


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