Question

Integrate the following w.r.t.x : `(1)/(2cosx + 3sinx)`

Answer 1

Let I = `int (1)/(2cosx + 3sinx)*dx`

= `int (1)/(3sinx + 2cosx)*dx`
Dividing numerator and denominator by
`sqrt(3^2 + 2^2) = sqrt(13)`, we get

I = `int ((1/sqrt(3)))/(3/sqrt(13) sinx + 2/sqrt(13) cosx)*dx`

Since, `(3/sqrt(13))^2 + (2/sqrt(13))^2 = (9)/(13) + (4)/(13)` = 1,

we take `(3)/sqrt(13) =cos oo, (2)/sqrt(13) = sin oo`

so that `oo = (2)/(3) and oo = tan^-1(2/3)`

∴ I = `(1)/sqrt(13) int (1)/(sin x + cosoo + cosx sin oo)*dx`

= `(1)/sqrt(13) int (1)/(sin(x + oo))*dx`

= `(1)/sqrt(13) int cosec (x + oo)*dx`

= `(1)/sqrt(13)log|tan|tan((x + oo)/2)| + c`

= `(1)/sqrt(13)log |tan ((x + tan^-1  2/3)/(2))| + c`.

Alternative Method

Let I = `int (1)/(2cosx + 3sinx)*dx`

Put `tan(x/2)` = t

∴ `x/(2) = tan^-1 t`

∴ x = 2tan^–1 t

∴ dx = `(2)/(1 + t^2)*dt`
and
sin x = `(2t)/(1 + t^2)` 
and
cos x = `(1 - t^2)/(1 + t^2)`

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

= `int (1 + t^2)/(2 - 2t^2 + 6t)*(2dt)/(1 + t^2)`

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

= `int (1)/(1 - (t^2 - 3t + 9/4) + 9/4)*dt`

= `int (1)/((sqrt(13)/2)^2 - (t - 3/2)^2)*dt`

= `(1)/(2 xx sqrt(13)/(2))log |(sqrt(13)/(2) + t - 3/2)/(sqrt(13)/(2) - t + 3/2)| + c`

= `(1)/sqrt(13)log|(sqrt(13) + 2t - 3)/(sqrt(13) - 2t + 3)| + c`

= `(1)/sqrt(13)log|(sqrt(13) + 2tan(x/2) - 3)/(sqrt(13) - 2tan(x/2) - 3)| + c`.