Question

The integral `int ((1 - 1/sqrt(3))(cosx - sinx))/((1 + 2/sqrt(3) sin2x))dx` is equal to ______.

Options

• `1/2log_e|(tan(x/2 + π/12))/(tan(x/2 + π/6))| + C`

• `1/2log_e|(tan(x/2 + π/6))/(tan(x/2 + π/3))| + C`

• `log_e|(tan(x/2 + π/12))/(tan(x/2 + π/12))| + C`

• `1/2log_e|(tan(x/2 - π/12))/(tan(x/2 - π/6))| + C`

Answer 1

The integral `int ((1 - 1/sqrt(3))(cosx - sinx))/((1 + 2/sqrt(3) sin2x))dx` is equal to `underlinebb(1/2log_e|(tan(x/2 + π/12))/(tan(x/2 + π/6))| + C`.

Explanation:

Given integral is I = `int ((1 - 1/sqrt(3))(cosx - sinx))/((1 + 2/sqrt(3) sin 2x))dx`

I = `sqrt(3)/2 int ((1 - 1/sqrt(3))(cosx - sinx))/((sqrt(3)/2 + sin 2x))dx`

I = `int ((sqrt(3)/2 - 1/2)(cos x - sin x))/(sin(π/3) + sin 2x)dx`

I = `int ((sqrt(3)/2 cos x - 1/2 cos x - sqrt(3)/2 sin x + 1/2 sin x))/(2sin(x + π/6)cos(x - π/6))dx`

I = `int ((cos(x - π/6) - sin(x + π/6)))/(2sin(x + π/6) cos(x - π/6))dx`

= `1/2(int (dx)/(sin(x + π/6)) - int (dx)/(cos(x - π/6)))`

= `1/2(int "cosec"(x + π/6)dx - int sec(x - π/6)dx)`

= `1/2 ln |(tan(x/2 + x/12))/(tan(x/2 + π/6))|`