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If 21+sinxdx=4cos(ax+b)+C then the value of a, b are respectively

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a
12,π4
b
1,π2
c
1,1
d
None of these

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detailed solution

Correct option is A

Let,I=∫21+sin⁡xdx=2∫sin⁡x2+cos⁡x2dx=2∫sin⁡π4+x2dx=−4cos⁡x2+π4+CBut,  I=−4cos⁡(ax+b)+Con comparing' we get a=12, b=π4


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