Let f n , x = ∫ n c o s n x d x , with f n , 0 = 0 . If the expression ∑ x = 1 89 f 1 , x simplifies to sin ⁡ a sin ⁡ b sin ⁡ c , then the value of b a c is (where a > b )

Step-by-step solution

f n , x = n s i n n x n + C As f n , 0 = C = 0 ⇒ f n , x = s i n n x Thus, ∑ x = 1 89 f 1 , x = sin ⁡ 1 + sin ⁡ 2 + . . . . . + sin ⁡ 89 = s i n 1 + 89 2 s i n 89 × 1 2 s i n 1 2 = s i n 45 s i n 89 2 s i n 1 2