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10 June, 23:57

Integrate sin^43x dx

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  1. 10 June, 23:59
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    Sstarica the question is ∫ sin4 3x dx

    substitute (cos x) ^2 = 1 - (sin x) ^2

    ∫ (1 - cos^2 (3x)) ^2 dx = ∫ (1-2 cos^2 u + cos^2 u) d u

    1/15 ∫ 5 (3) sin^4 (3x) d x

    = - cos^5 (3x) / 15 + C
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