please help me with this integration
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your answer is wrong
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Hello!
✓ Integral cos²x sin³x dx :
= ∫ cos²x sin²x sinx dx
= ∫ (¹/₂ + ¹/₂ cos 2x) (¹/₂ - ¹/₂ cos 2x) sinx dx
= ∫ (¹/₄ - ¹/₄ cos² 2x) sinx dx
= ∫ [ ¹/₄ - (¹/₈ + ¹/₈ cos 4x)] sin x dx
= ∫ [ ¹/₄ - ¹/₈ - ¹/₈ cos 4x] sin x dx
= ∫ (-⁶/₈ - ¹/₈ cos 4x ) sin x dx
= ∫ (-⁶/₈ sin x - ¹/₈ cos 4x sin x ) dx
= ∫ (-⁶/₈ sin x -¹/₁₆ ( sin 5x - sin 3x) dx
= ∫ (-⁶/₈ sin x -¹/₁₆ sin 5x - ¹/₁₆ sin 3x) dx
= ⁶/₈ cos x + ¹/₈₀ cos 5x - ¹/₄₈ sin 3x + c
Cheers!
✓ Integral cos²x sin³x dx :
= ∫ cos²x sin²x sinx dx
= ∫ (¹/₂ + ¹/₂ cos 2x) (¹/₂ - ¹/₂ cos 2x) sinx dx
= ∫ (¹/₄ - ¹/₄ cos² 2x) sinx dx
= ∫ [ ¹/₄ - (¹/₈ + ¹/₈ cos 4x)] sin x dx
= ∫ [ ¹/₄ - ¹/₈ - ¹/₈ cos 4x] sin x dx
= ∫ (-⁶/₈ - ¹/₈ cos 4x ) sin x dx
= ∫ (-⁶/₈ sin x - ¹/₈ cos 4x sin x ) dx
= ∫ (-⁶/₈ sin x -¹/₁₆ ( sin 5x - sin 3x) dx
= ∫ (-⁶/₈ sin x -¹/₁₆ sin 5x - ¹/₁₆ sin 3x) dx
= ⁶/₈ cos x + ¹/₈₀ cos 5x - ¹/₄₈ sin 3x + c
Cheers!
answer is wrong
What's the correct answer, Then?
(minus one upon cos cube x) plus (one upon 5 cos raise 5 x )+ c
It's more simplified form brother! :)
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