Question

`int 1/(2sin2x - 3)dx` = ______ 

विकल्प

• `1/5tan^-1((3tanx - 2)/sqrt5) + c`

• `-1/5tan^-1((3tanx - 2)/sqrt5) + c`

• `1/sqrt5tan^-1((3tanx - 2)/sqrt5) + c`

• `-1/sqrt5tan^-1((3tanx - 2)/sqrt5) + c`

Answer 1

`int 1/(2sin2x - 3)dx` = `underline(-1/sqrt5tan^-1((3tanx - 2)/sqrt5) + c)`

Explanation:

Let I = `int 1/(2sin2x - 3)dx`

Put tan x = t ⇒ x = tan^-1 t

⇒ dx = `(dt)/(1 + t^2)` and sin 2x = `(2t)/(1 + t^2)`

∴ I = `int(((dt)/(1 + t^2)))/(2((2t)/(1 + t^2))- 3) = int(dt)/(4t - 3 - 3t^2)`

= `int dt/(-3t^2 + 4t - 3) = -1/3int dt/(t^2 - 4/3t + 1)`

= `-1/3intdt/(t^2 - 4/3t + 4/9 - 4/6 + 1)`

= `-1/3intdt/((t - 2/3)^2 + 5/9)`

= `-1/3intdt/((t - 2/3)^2 + (sqrt5/3)^2)`

= `-1/3 . 1/((sqrt5/3)) tan^-1 ((t - 2/3)/(sqrt5/3)) + c`

= `(-1)/sqrt5 tan^-1 ((3t - 2)/sqrt5) + c`

∴ I = `-1/sqrt5 tan^-1 ((3tanx - 2)/sqrt5) + c`