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Question

cosec(2x+5)dx is equal to
(where C is constant of integration)

A
ln|cosec(2x+5)+cot(2x+5)|2+C
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B
ln|cosec(2x+5)+cot(2x+5)|5+C
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C
ln|cosec(2x+5)+cot(2x+5)|5+C
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D
ln|cosec(2x+5)+cot(2x+5)|2+C
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Solution

The correct option is D ln|cosec(2x+5)+cot(2x+5)|2+C
cosec(2x+5)dx=ln|cosec(2x+5)+cot(2x+5)|2+C
(cosec(ax+b)dx=ln|cosec(ax+b)+cot(ax+b)|a+C,a0)

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