Question:

\pi 3 \pi sin 2 ( \pi - x)dx 2 (sin x) 2 dx 2 2 (3) I = \int 3 3 = \int 3 3 \pi \pi 0 0 (sin x) 2 x + (cos x) 2 sin 2 ( - x) + cos 2 ( - x) 2 2 3 3 \pi 2 (sin x) 2 + (cos x) 2 \pi \Rightarrow Adding 2l = \int dx = 3 3 2 0 (sin x) 2 + (cos x) 2 \pi \pi \pi ( - x) sin x cos x 2 x sin x cos x 2 2 I0 = \int dx = \int dx 4 4 4 0 sin x + cos 4 x 0 (sin x) + (cos x) \pi \pi (sin x) cos x Adding, 2 2 2I0 = \int dx 0 4 4 (sin x+cos x) \pi 2 tan x(sec x)dx \pi 2 \Rightarrow I0 = \int 4 0 4 1+tan x put tan x = t \Rightarrow 2 tan x sec xdx = dt 2 2 dt \infty \infty \pi 2 \pi ∣ \pi \pi -1 \Rightarrow I0 = \int = (tan t)∣ = ( - 0) 4 2 8 ∣ 8 2 0 (1 + t ) 0 2 \pi \Rightarrow I0 = 16 \pi