Please show the process for integration of tanx23secx23 tan

Solution

integral (-tan^3(x/2)+sec^3(x/2) tan^3(x/2)) dx Integrate the sum term by term and factor out constants: = integral tan^3(x/2) sec^3(x/2) dx- integral tan^3(x/2) dx For the integrand tan^3(x/2) sec^3(x/2), substitute u = x/2 and du = 1/2 dx: = 2 integral tan^3(u) sec^3(u) du- integral tan^3(x/2) dx For the integrand tan^3(u) sec^3(u), use the trigonometric identity tan^2(u) = sec^2(u)-1: = 2 integral tan(u) sec^3(u) (sec^2(u)-1) du- integral tan^3(x/2) dx For the integrand tan(u) sec^3(u) (sec^2(u)-1), substitute s = sec(u) and ds = tan(u) sec(u) du: = 2 integral s^2 (s^2-1) ds- integral tan^3(x/2) dx Expanding the integrand s^2 (s^2-1) gives s^4-s^2: = 2 integral (s^4-s^2) ds- integral tan^3(x/2) dx Integrate the sum term by term and factor out constants: = 2 integral s^4 ds-2 integral s^2 ds- integral tan^3(x/2) dx The integral of s^2 is s^3/3: = 2 integral s^4 ds-(2 s^3)/3- integral tan^3(x/2) dx The integral of s^4 is s^5/5: = (2 s^5)/5-(2 s^3)/3- integral tan^3(x/2) dx For the integrand tan^3(x/2), substitute p = x/2 and dp = 1/2 dx: = -2 integral tan^3(p) dp+(2 s^5)/5-(2 s^3)/3 Use the reduction formula, integral tan^m(p) dp = (tan^(m-1)(p))/(m-1) - integral tan^(-2+m)(p) dp, where m = 3: = -tan^2(p)+2 integral tan(p) dp+(2 s^5)/5-(2 s^3)/3 The integral of tan(p) is -log(cos(p)): = -tan^2(p)-2 log(cos(p))+(2 s^5)/5-(2 s^3)/3+constant Substitute back for p = x/2: = (2 s^5)/5-(2 s^3)/3-tan^2(x/2)-2 log(cos(x/2))+constant Substitute back for s = sec(u): = (2 sec^5(u))/5-(2 sec^3(u))/3-tan^2(x/2)-2 log(cos(x/2))+constant Substitute back for u = x/2: = -tan^2(x/2)+2/5 sec^5(x/2)-2/3 sec^3(x/2)-2 log(cos(x/2))+constant Which is equivalent for restricted x values to: = 2/5 sec^5(x/2)-2/3 sec^3(x/2)-sec^2(x/2)-2 log(cos(x/2))+constant