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Question

dxcosx+3sinx equals

A
logtan(x2+π12)+c
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B
logtan(x2π12)+c
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C
12logtan(x2+π12)+c
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D
12logtan(x2π12)+c
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Solution

The correct option is C 12logtan(x2+π12)+c
I=dxcosx+3sinxI=dx2(12cosx+32sinx)I=12dxsinπ6cosx+cosπ6sinxI=12dxsin(π6+x)I=12csc(π6+x)dxI=12log{csc(π6+x)cot(π6+x)}+cI=12log{1cos(π6+x)sin(π6+x)}+cI=12log{2sin2(π12+x2)2sin(π12+x2)cos(π12+x2)}+cI=12log{tan(π12+x2)}+c

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