The anti – derivative of the function (3x + 4) |sin x|, when 0<x<π, is given by
A
3sinx−(3x+4)cosx+c
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B
3sinx+(3x+4)cosx+c
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C
−3sinx−(3x+4)cosx+c
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D
None of these
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Solution
The correct option is A3sinx−(3x+4)cosx+c In the interval (0,π), sin x is positive, therefore, (3x+4)|sinx|=(3x+4)sinx. ∴The antiderivative of (3x+4)|sinx| is =∫(3x+4)sinxdx=−(3x+4)cosx+∫3cosxdx=−(3x+4)cosx+3sinx+c