Question

Evaluate the following:

`int_0^(pi/2) ("d"x)/(1 + 5cos^2x)`

Answer 1

Let I = `int_0^(pi/2) ("d"x)/(1 + 5cos^2x)`

Dividing both Numerator and Denominator by cos²x.

= `int_0^(pi/2) (1/(cos^2x))/(1/(cos^2x) + 5)  "d"x`

= `int_0^(pi/2) (sec^2x)/(sec^2x + 5)  "d"x`

= `int_0^(pi/2) (sec^2x)/(1 + tan^2x + 5)  "d"x`

= `int_0^(pi/2) (sec^2x)/(tan^2x+ 6)  "d"x`  ........(1)

Consider, t = tan x   ........(2)

DIfferentiate with respect to 'x'

dt = sec²x dx .......(3)

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

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

(1) ⇒ = `int_0^oo "dt"/("t"^2 + 6)`

= `int_0^oo "dt"/("t"^2 + (sqrt(6))^2`  ........`{{:(int ("d"x)/("a"^2+ x^2) = 1/"a"tan^-1 (x/"a")), (tan-1 (oo) = pi/2):}}`

= `[1/sqrt(6) tan^-1  "t"/sqrt(6)]_0^oo`

= `1/sqrt(6) tan^-1 (oo) - 0`

= `1/sqrt(6)(pi/2)`

= `pi/(2sqrt(6))`