The integral `int(sec^2x)/((secx+tanx)^(9/2))dx` equals (for some arbitrary constant `K)dot` `-1/((secx+tanx)^((11)/2)){1/(11)-1/7(secx+tanx)^2}+K` `1/((secx+tanx)^(1/(11))){1/(11)-1/7(secx+tanx)^2}+K` `-1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K` `1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K`

Options

A. `(-1)/((secx + tanx)^(11//2)){(1)/(11)-(1)/(7)(secx + tanx)^(2)}+K`
B. `(1)/((secx + tanx)^(11//2)){(1)/(11)-(1)/(7)(secx + tanx)^(2)}+K`
C. `(-1)/((secx + tanx)^(11//2)){(1)/(11)+(1)/(7)(secx + tanx)^(2)}+K`
D. `(1)/((secx + tanx)^(11//2)){(1)/(11)+(1)/(7)(secx + tanx)^(2)}+K`

Answer

C

Explanation

Plan Integration by Substitution
`i.e." "I=intf{g(x)}*g'(x)dx`
`Put" "g(x)=t rArr g'(x)dx = dt`
`therefore" "I = int f(t)dt`
Description of Situation Generally, students gets consfused after substitution, i.e. sec x + tan x = t. Now, for sec x, we should use
`" "sec^(2)x-tan^(2)x=1`
`rArr" "(sec x-tan x)(sec x+tan x)=1`
`rArr" "sec x - tan x = (1)/(t)`
`"here"," "I=int(sec^(2)dx)/((sec x + tan x)^(9//2))`
Put sec x + tan x = t
`rArr (sec x tan x + sec^(2)x)dx = dt`
`rArr" "sec x*"t dx"=dt rArr sec "x dx"=(dt)/(t)`
`therefore" "sec x-tan x = (1)/(t) rArr sec x =(1)/(2)(t+(1)/(t))`
`therefore" "I=int(sec x*sec "x dx")/((sec x + tan x)^(9//2))`
`rArr I = int((1)/(2)(t+(1)/(t))*(dt)/(t))/(t^(9//2))=(1)/(2)int((1)/(t^(9//2))+(1)/(t^(13//2)))dt`
`=-(1)/(2){(2)/(7t^(7//2))+(2)/(11t^(11//2))}+K`
`=-[(1)/(7(sec x + tan x)^(7//2))+(1)/(11(sec x + tan x)^(11//2))]+K`
`=(-1)/((sec x + tan x)^(11//2)){(1)/(11)+(1)/(7)(sec x + tan x)^(2)}+K`