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\( \int \sin ^{2} x \cdot \cos ^{4} x d x \) |
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Answer» ∫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) |
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