Question

Evaluate the following:

`int_0^(pi/2) ("e"^(-tanx))/(cos^6x)  "d"x`

Answer 1

= `int_0^(pi/2) "e"^(-tanx) sec^6  x  "d"x`

= `int_0^(pi/2) "e"^(-tanx) sec^4x sec^2x  "d"x`  ..........(1)

Put t = tan x    .........(2)

Differentiate with respect to 'x'

dt = sec²x dx   .........(3)

x	0	`pi/2`
t	0	`oo`

Sbstute (2) and (3) in (1), we get

(1) ⇒ = `int_0^oo "e"^(-"t")(sec^2x)^2  "dt"`  ........`(because  sec^2x = 1+ tan^2x = 1 + "t"^2)`

= `int_0^o "e"^(-"t") (1 + "t"^2)^2  "dt"`

= `int_0^oo "e"^(-"t") (1 + "t"^4 + 2"t"^2)  "dt"`

= `int_0^oo "e"^(-"t") "dt" + int_0^oo "e"^(-"t")"t"^4  "dt"  2int_0^oo "e"^(-"t")"t"^2 "dt"`

By ussing Gamma Integral,

= `(- "e"^(-"t"))_0^oo + (4!)/((1)^(4 + 1)) + 2((2!))/((1)^(2 + 1))`

= `-"e"^(- oo) + "e"^0 + 4! + 2(2!)`

= 0 + 1 + 24 + 4

= 29