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$If I_{n} = \int \frac{\cos n x}{\sin x} d x, then I_{10} - I_{8} is equal to$

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−29cos11x+C

−211cos9x+C

29cos9x+C

−29cos9x+C

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detailed solution

Correct option is C

I10=∫cos10xsinxdxcos⁡10x−cos⁡8x=−2sin⁡9xsin⁡x⇒cos⁡10x−cos⁡8xsin⁡x=−2sin⁡9x∴∫cos⁡10xsin⁡x−cos⁡8xsin⁡xdx=∫−2sin⁡9xdx⇒I10−I8=−2(−cos⁡9x)9⇒I10−I8=29cos⁡9x+C

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If In=∫cos⁡nxsin⁡xdx, then I10−I8 is equal to

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$If I_{n} = \int \frac{\cos n x}{\sin x} d x, then I_{10} - I_{8} is equal to$