`int(sec^(8)x)/(cosecx) dx =`

Options

A. `(sec^(8)x)/(8) + c`
B. `(sec^(7) x)/(7) + c`
C. `(sec^(6) x)/(6) + c`
D. `(sec^(9)x)/(9) + c`

Answer

b

Explanation

Let `I = int(sec^(8)x)/(cosecx) dx = int(sinx)/(cos^(8))dx`
`[ :' (1)/(cosecx) = sinx " and " sec x = (1)/(cosx)]`
`= int tanx . sec^(7)xdx`
`rArr I = intsec^(6)x. secx tanx dx`
On putting `secx = t`
`rArr sec x . tan x dx = dt`,
we get, `I = intt^(6) dt = (t^(7))/(7) +c`
`= (sec^(7)x)/(7) + c` `[ :' t = sec x]`