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Question

Solve cotxcos4x+sin4xdx=......+c

A
12logcot2x+cot4x+1
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B
12logcot2x+cot4x+1
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C
12logtan2x+tan4x+1
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D
12logcotx+cot4x+1
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Solution

The correct option is B 12logcot2x+cot4x+1
I=cotxcos4x+sin4xdx

I=cotx sin4x(cos4xsin4x+1)dx

I=cotxsin2xcot4x+1dx

I=cotxcosec2xcot4x+1dx

Now take cot2x=t

2cotxcosec2xdx=dt
cotxcosec2xdx=12dt

I=121t2+1dt=12logt+t2+1+c

I=12logcot2x+cot4x+1+c

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