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a
cotx22+logsinx+C
b
−3cot2xcosec2x+C
c
3cot2xcosec2x+C
d
−cotx22+logcosecx+C
answer is D.
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Detailed Solution
I=∫cot3xdx=∫cot2xcotxdx I=∫cosec2x−1cotxdx=∫cotxcosec2xdx−∫cotxdx Let I1=∫cotxcosec2xdxcotx=t→−cosec2xdx=dtI1=∫tdt=−t22I=−(cotx)22−log|sinx|+C