Question

integrate the function [2cosx - 3sinx]/[6cosx + 4sinx].dx

Answer 1

$\bf{I=\int{\frac{2cosx-3sinx}{6cosx+4sinx}}\,dx}\\\\=\bf{\int{\frac{2cosx-3sinx}{2(3cosx+2sinx)}}\,dx}$

Let (3cosx + 2sinx) = f(x)

differentiate both sides,

(-3sinx + 2cosx) = f'(x)

(2cosx - 3sinx) = f'(x) , put it in I.

so, I=$\bf{\int{\frac{f'(x)}{2f(x)}}\,dx}$

= $\frac{1}{2}\bf{log_e|f(x)|}+C$

now, put f(x) = (3cosx + 2sinx)

then, I=$\frac{1}{2}\bf{log_e|3cosx+2sinx|}+C$