Correct Answer - C
Let `I_(9)=int tan^(9)xdx=int tan^(7)x(sec^(2)x-1)dx`
`"i.e., "I_(9)=(tan^(8)x)/(8)-I_(7)`
`"Similarly, "I_(7)=(tan^(6)x)/(6)-I_(5)`
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`I_(3)=(tan^(2)x)/(2)-I_(1)=(tan^(2)x)/(2)-int tanxdx`
`=(tan^(2)x)/(2)+log|cos x|`
`therefore I_(9)=" (a polynomial of degree 8 in tan x)"+ log |cos x|+C`