Take the integral: integral csc(x+1) csc(x+2) dx Write csc(x+1) csc(x+2) as 2/(cos(1)-cos(2 x+3)): = integral 2/(cos(1)-cos(2 x+3)) dx Factor out constants: = 2 integral 1/(cos(1)-cos(2 x+3)) dx For the integrand 1/(cos(1)-cos(2 x+3)), substitute u = 2 x+3 and du = 2 dx: = integral 1/(cos(1)-cos(u)) du For the integrand 1/(cos(1)-cos(u)), substitute s = tan(u/2) and ds = 1/2 sec^2(u/2) du. Then transform the integrand using the substitutions sin(u) = (2 s)/(s^2+1), cos(u) = (1-s^2)/(s^2+1) and du = (2 ds)/(s^2+1): = integral 2/((s^2+1) (cos(1)-(1-s^2)/(s^2+1))) ds Simplify the integrand 2/((s^2+1) (cos(1)-(1-s^2)/(s^2+1))) to get 2/(s^2+s^2 cos(1)-1+cos(1)): = integral 2/(s^2+s^2 cos(1)-1+cos(1)) ds Factor out constants: = 2 integral 1/(s^2+s^2 cos(1)-1+cos(1)) ds Factor cos(1)-1 from the denominator: = 2 integral 1/((cos(1)-1) ((s^2 (1+cos(1)))/(cos(1)-1)+1)) ds Factor out constants: = 2/(cos(1)-1) integral 1/((s^2 (1+cos(1)))/(cos(1)-1)+1) ds For the integrand 1/((s^2 (1+cos(1)))/(cos(1)-1)+1), substitute p = s cot(1/2) and dp = cot(1/2) ds: = (2 tan(1/2))/(cos(1)-1) integral 1/(1-p^2) dp The integral of 1/(1-p^2) is tanh^(-1)(p): = (2 tan(1/2) tanh^(-1)(p))/(cos(1)-1)+constant Substitute back for p = s cot(1/2): = csc(1/2) sec(1/2) (-tanh^(-1)(s cot(1/2)))+constant Substitute back for s = tan(u/2): = csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(u/2)))+constant Substitute back for u = 2 x+3: = csc(1/2) sec(1/2) (-tanh^(-1)(cot(1/2) tan(x+3/2)))+constant Which is equivalent for restricted x values to: Answer: | | = csc(1) (log(sin(x+1))-log(sin(x+2)))+constant