Question

`int_(pi/18)^((4pi)/9) (2 sqrt(sin x))/(sqrt (sin x) + sqrt(cos x))` dx = ?

Options

• `(7pi)/36`

• `(5pi)/36`

• `(7pi)/18`

• `(5pi)/18`

Answer 1

`(7pi)/18`

Explanation:

Let I = `int_(pi/18)^((4pi)/9) (2 sqrt(sin x))/(sqrt (sin x) + sqrt(cos x))` dx

`= 2 int_(pi/18)^((4pi)/9) (sqrt(sin x))/(sqrt (sin x) + sqrt(cos x))` dx     ...(i)

`= 2 int_(pi/18)^((4pi)/9) sqrt(sin ((4pi)/9 + pi/18 - x))/((sqrt(sin (4pi)/9 + pi/8 - x)) + sqrt(cos ((4pi)/ 9 + pi/8 - x)))`   ....`(because int_"a"^"b" "f"(x) "dx" = int_"a"^"b" "f"("a + b" - x) "dx")`

`=> "I" = 2 int_(pi/18)^((4pi)/9) sqrt(cos x)/(sqrt(cos x) + sqrt(sin x))`dx    ....(ii)

On adding Eqs. (i) and (ii), we get

2I = `2 int_(pi/18)^((4pi)/9) "dx" = 2[x]_(pi/18)^((4pi)/9)`

`= 2[(4pi)/9 - pi/18]`

`= 2((7pi)/18)`

`=> "I" = (7pi)/18`