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Q.

In=∫cotn⁡xdx then I10 is equal to

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a

cot9x9−I8+C

b

−cot9x9−I8+C

c

−cot11x11−I9+C

d

cot11x11−I9+C

answer is B.

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Detailed Solution

I10+I8=∫cot10xdx+∫cot8xdx=∫cot8⁡xcot2⁡x+1dx=∫cot8⁡xcosec2⁡xdxput    cot⁡x=t=∫t8(−dt)=−t99=−cot9⁡x9I10=−cot9⁡x9−I8+C
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