Let's find a reduction formula for the indefinite integral In = sec"(x) dac . If n > 2 then we can rewrite this integral as In = (1+ tan?(x)) sec"-2(2) da = In-2 + / tan?(x) sec"-2(x) da. Now use integration by parts to get tan?(x) secn -2(x) da = u' da = [uv] - uvdx with u = tan(x) and v = tan(x) secn-2(x) . This gives . [uv] = (tan(x)*sec(x)^(n-2)/(n-2)- and . uv = (sec(x)^2*sec(x^(n-2)))/(n. Hence, without further integration we have the reduction formula In = _ sec-2(x) tan(x) + g(n) In-2 where g ( n) = (n-2)/(n-1) Use this to evaluate 14 = (1/3)*tan(x)*sec(x)^2+(2/3 +C Recall the Maple syntax for sin (a ) is sin(x)^2