Question

\( \int \sin ^{2} x \cdot \cos ^{4} x d x \)

Answer 1

∫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)