InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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`A. `-(1)/((secx+tanx)^(11//2)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`B. `(1)/((secx+tanx)^(1//11)){(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» Correct Answer - C `I=int(sec^(2)x)/((secx+tanx)^(9//2))dx` Let ` sec x +tanx=t` ` or sec x -tan x=1//t` Now, `(secx tanx+sec^(2)x)dx =dt` `or secx(secx+tanx)dx=dt` ` or secx dx=(dt)/(t),(1)/(2)(t+(1)/(t))=sec x` ` :. I=(1)/(2)int ((t+(1)/(t)))/(t^(9//2))(dt)/(t)` `=(1)/(2)int(t^(-9//2)+t^(-13//2))dt` `=(1)/(2)[(t^(-9//2+1))/(-(9)/(2)+1)+(t^(-13//2+1))/(-(13)/(2)+1)]+K` `=(1)/(2)[(t^(-7//2))/(-(7)/(2))+(t^(-11//2))/(-(11)/(2))]+K` `= -(1)/(7) t^(-7//2)-(1)/(11)t^(-11//2)+K` `= -(1)/(7) (1)/(t^(7//2))-(1)/(11)(1)/(t^(11//2))+K` `= -(1)/(t^(11//2))((1)/(11)+(t^(2))/(7))+K` `= -(1)/((secx+tan x)^(11//2)){(1)/(11)+(1)/(7)(sec x+tanx)^(2)}+K` |
|