If the value of the integral ∫ 0 π 2 c o t θ 2 1 - c o s 4 θ d θ is equal to λ , then the value of 3 λ is equal to

Step-by-step solution

Let I = ∫ 0 π 2 c o t θ 2 1 - c o s θ 1 + c o s θ + c o s 2 θ + c o s 3 θ d θ So, I = ∫ 0 π 2 c o s θ 2 s i n θ 2 · 2 s i n 2 θ 2 1 + c o s θ + cos 2 ⁡ θ + cos 3 ⁡ θ d θ = ∫ 0 π 2 sin ⁡ θ 1 + cos θ + cos 2 θ + cos 3 θ d θ Let cos ⁡ θ = t ⇒ I = ∫ 0 1 1 + t + t 2 + t 3 d t = 1 + 1 2 + 1 3 + 1 4 = 25 12 Hence, λ = 25 12 ⇒ 3 λ = 25 4 = 6