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Question

In the integral
cos8x+1cot2xtan2xdx=Acos8x+k is an arbitrary constant, then A is equal to

A
116
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B
116
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C
18
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D
19
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Solution

The correct option is A 116
cos8x+1tan2xcot2xcos8x+1Sm2xcos2xcos2xsin2xdx=2sinx.cosx.(cos8x+1)2(sin2xcos22x)dx=sin4x(cos8x+1)2cos4xdx=12tan4x[cos8x+1]dx4x=tdx=14dt=12×4tant(cos2t+1)dt=18sintxcost(2cos2t1+1)dt=18sin2tdt=18[cos2t2]+c=116cos2t+k=Acos8x+k=116cos8x+kA=116

Hence, this is the answer.

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