$$In=$$∫cotn⁡xdx then $$I10$$ is equal to

Correct option is 2−cot9x9−I8+C

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$I_{n} = \int \cot^{n} x d x then I_{10} is equal to$

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cot9x9−I8+C

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detailed solution

Correct option is B

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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In=∫cotn⁡xdx then I10 is equal to

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$I_{n} = \int \cot^{n} x d x then I_{10} is equal to$