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Q.

∫cot3xdx=

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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=∫cot3⁡xdx=∫cot2⁡xcot⁡xdx I=∫cosec2⁡x−1cot⁡xdx=∫cot⁡xcosec2⁡xdx−∫cot⁡xdx Let I1=∫cot⁡xcosec2⁡xdxcot⁡x=t→−cosec2⁡xdx=dtI1=∫tdt=−t22I=−(cot⁡x)22−log⁡|sin⁡x|+C
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