Let's find a reduction formula for the indefinite integral In = sec" (x) da . If n > 2 then we can rewrite this integral as In = (1 +tan?(x)) sec"-2(x) dx = In-2 + /tan?(x) sec"-2(x) da. Now use integration by parts to get tan?(x) sec -2(x) dx = wv' dx = [uv] - u'vdz with u = tan(x) and v = tan(x) sec-2 (x). This gives . [uv] and . uv = Hence, without further integration we have the reduction formula In = 1 n-1 sec-2(x) tan(x) + g(n)In-2 where g( n) = Use this to evaluate IA = A +C