Question

Evaluate the following:

`int_0^(pi/2) ("d"x)/(5 + 4sin^2x)`

Answer 1

Let I = `int_0^(pi/2) ("d"x)/(5 + 4sin^2x)`

Dividing both Numerator and Denominator by cos²x

= `int_0^(pi/2) (1/(cos^2x))/(5/(cos^2x)+ (4sin^2x)/(cos^2x))  "d"x`

= `int_0^(pi/2) (sec^2x)/(5sec^2x + 4tan^2x)  d"x`

= `int_0^(pi/2) (sc^2x)/(5(1  tan^2x) + 4tan^2x)`

= `int_0^(pi/2) (sec^2 x)/(5 + 9tan^2x)  "dx`  ..........(1)

Considerr, t = tan x    ........(2)

Diferentite with rect t '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"/(5 + 9"t"^2)`

= `1/9 int_0^oo  "dt"/(5/9+ "t"^2)`

= `1/9 int_0^oo  "dt"/((sqrt(5)/3)^2 + "t"^2)`

= `1/9[1/(sqrt(5)/3) tan^-1  "t"/(sqrt5/3)]_0^oo`

= `1/9 [3/sqrt(5) (pi/2) - 0]`

 `1/9[(3pi)/(2sqrt(5))]`

 `pi/(6sqrt5)`