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If dxsinxcosx=log|f(x)|+C then

 

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a
f(x)=sin⁡x+cos⁡x
b
f(x)=tan⁡x
c
f(x)=sec2⁡x
d
none of these

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detailed solution

Correct option is B

∫dxsin⁡xcos⁡x=∫sin2⁡x+cos2⁡xsin⁡xcos⁡xdx=∫(tan⁡x+cot⁡x)dx=−log⁡|cos⁡x|+log⁡|sin⁡x|+C=log⁡|tan⁡x|+C. Thus f(x)=tan⁡x.


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