If a = ∫ 0 1 c o s sin ⁡ x sec ⁡ x d x , then the value of a 2 + c o s 2 sin ⁡ 1 is equal to

Step-by-step solution

a = ∫ 0 1 cos ⁡ sin ⁡ x cos ⁡ x d x Let sin ⁡ x = t ⇒ cos ⁡ x d x = d t ∴ a = ∫ 0 sin ⁡ 1 cos ⁡ t d t = ( sin ⁡ t ) 0 sin ⁡ 1 = s i n sin ⁡ 1 - sin ⁡ 0 = s i n sin ⁡ 1 ∴ a 2 + c o s 2 sin ⁡ 1 = sin 2 ⁡ θ + cos 2 ⁡ θ = 1 ;   θ = sin ⁡ 1