The option(s) with the value of a and L that satisfy the following equation is(are) ∫ 0 4 π e t sin 6 ⁡ a t + cos 4 ⁡ a t d t ∫ 0 π sin 6 ⁡ a t + cos 4 ⁡ a t d t = L ?

Step-by-step solution

l = ∫ 0 4 π e t s i n 6 2 t + c o s 4 2 t d t = l 1 + l 2 + l 3 + l 4 W h e r e l 1 = ∫ 0 π e t sin 6 ⁡ 2 t + cos 4 ⁡ 2 t d t a n d l 2 = ∫ π 2 π e t sin 6 ⁡ 2 t + c o s 4 2 t d t a n d l 3 = ∫ 2 π 3 π e t sin 6 ⁡ 2 t + cos 4 ⁡ 2 t d t a n d l 4 = ∫ 3 π 4 π e t s i n 6 2 t + c o s 4 2 t d t t = x + π d t = d x l 2 = ∫ 0 π e x + π s i n 6 2 x + c o s 4 2 x d x = e π l 1 t = 2 π + x l 3 = e 2 π l 1 l 4 = e 3 π l 1 l = l 1 1 + e π + e 2 π + e 3 π l l 1 = e 4 π - 1 e π - 1 The same for a = 4.