#I= int(tan(2x-3) - cot(3x))dx#
#I =inttan(2x-3)dx - int(cot(3x)dx#
#I =int sin(2x-3)/cos(2x-3)dx - int cos(3x)/sin(3x)dx#
#I=I_1 - I_2 #
Let us name the two integrals as #I_1# and #I_2#
and solve them separately.
#I_1 = int sin(2x-3)/cos(2x-3) dx#
Let us solve this first
Let #u=cos(2x-3)#
Diffrentiating with respect to #x# using chain rule.
#(du)/dx = -sin(2x-3)d/dx(2x-3)#
#(du)/dx = -sin(2x-3)(2)#
#(du)/dx = -2sin(2x-3)#
#du = -2sin(2x-3)dx#
#(du)/-2 = sin(2x-3)dx#
#I_1 = int 1/u ((du)/-2)#
#I_1 = -1/2int (du)/u#
#I_1 = -1/2ln(u)#
#I_1 = -1/2 ln(cos(2x-3))#
#I_2 = intcos(3x)/sin(3x) dx#
Let #u=sin(3x)#
Differentiating with respect to #x# using chain rule.
#(du)/dx = cos(3x)d/dx(3x)#
#(du)/dx = cos(3x)3#
#(du)/dx =3cos(3x)#
#(du)/3 = cos(3x)dx#
#I_2 = int1/u ((du)/3)#
#I_2 = 1/3 int (du)/u#
#I_2 = 1/3 ln(u)#
#I_2 = 1/3 ln(sin(3x))#
We had got till the step #I=I_1-I_2#
# -1/2ln(cos(2x-3)) - 1/3ln(sin(3x)) + C# Answer
