`int (sec^(2) x)/((sec x+ tan x)^(9//2)) dx` का मान ज्ञात कीजिए

Explanation

माना `I=int (sec^(2) x)/((sec x+tan x)^(9//2)) dx`
माना `t=sec x+tan x`
`dt=(sec x tan x+sec^(2) x) dx`
`dt=sec x(tan x+sec x) dx`
`(dt)/(sec x+tan x) =sec x dx`
`(dt)/t=sec x dx`
और `sec x-tan x=1/t" "[ :' sec^(2) x-tan^(2) x=1 से]`
`:. sec x=1/2 (t+1/t)`
अब पुनः `I=int (sec x.sec x)/((sec x+tan x)^(9//2)) dx`
`=int (1/2 (t+1/t).(dt)/t)/t^(9//2)`
`=1/2 int (dt)/t^(9//2)+(dt)/t^(13//2)`
`=1/2 int t^(-9//2) dt+t^(-13//2) dt`
`=1/2 [t^(-9/2+1)/(-9/2+1)+t^(-13/2+1)/(-13/2+1)]`
`=1/2 [- 2/(7t^(7//2))-2/(11t^(11/2))]`
`=-[(1)/(7(sec x+tan x)^(7//2))+1/(11(sec x+tan x)^(11//2))]`.