Evaluate:
#int (1-cos(x/3))/sin(x/2)dx#
Substitute #t= x/6#
#int (1-cos(x/3))/sin(x/2)dx = 6 int (1-cos 2t)/sin(3t) dt#
use the trigonometric identities:
#1-cos 2t = 2sin^2t#
#sin 3t = 3sint-4sin^3t#
to have:
#int (1-cos(x/3))/sin(x/2)dx = 12 int sin^2t/( 3sint-4sin^3t) dt#
simplifying:
#int (1-cos(x/3))/sin(x/2)dx = 12 int sint/( 3-4sin^2t) dt#
Now write the denominator as:
#3-4sin^2t = 4 -4sin^2t -1 = 4(1-sin^2t) -1 = 4cos^2t-1#
so:
#int (1-cos(x/3))/sin(x/2)dx = 12 int sint/( 4cos^2t-1) dt#
and substituting #u=cost# #du= -sint dt#
#int (1-cos(x/3))/sin(x/2)dx = -12 int (du)/( 4u^2-1) dt#
decompose the rational function in partial fractions:
#1/(4u^2-1) = 1/((2u-1)(2u+1)) = A/(2u-1)+B/(2u+1)#
#A(2u+1)+B(2u-1) = 1#
#2(A+B)u + (A-B) = 1#
#{(A+B = 0),(A-B = 1):}#
#{(A=1/2),(B=-1/2):}#
#int (1-cos(x/3))/sin(x/2)dx = -6 int (du)/(2u-1) +6int (du)/(2u+1)#
#int (1-cos(x/3))/sin(x/2)dx = -3 ln abs(2u-1) +3 ln abs (2u+1) +C #
undoing the substitution and using the properties of logarithms:
#int (1-cos(x/3))/sin(x/2)dx = 3 ln abs((2cos (x/6)+1)/ (2cos(x/6)-1) )+C #
