∫sin2x cos4x dx
= 1/4 ∫(2 sin x cos x)2 cos2x dx
= 1/8 ∫sin22x x 2cos2x dx (∵ 2 sinθ cosθ = sin2θ)
= 1/8 ∫sin22x(1 + cos2x) dx (∵ cos2θ = 2cos2θ - 1 ⇒ 2 cos2θ = cos2θ + 1)
= 1/8 ∫sin22x cos2x dx + 1/8 ∫sin22x dx
= 1/8 ∫t2 dt/2 + 1/8 ∫\(\frac{1-cos4x}{2}dx\)
(By taking sin 2x = t ⇒ cos2x dx = dt/2 And sin2θ \(=\frac{1-cos2\theta}{2})\)
= 1/16 x t3/3 + 1/16 (x - sin4x/4) + c
= 1/48 sin32x + x/16 - 1/64 sin4x + c (By putting t = sin 2x)