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Trustted

  • 3 years ago

integrate sin^3X

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  1. Jonask
    • 3 years ago
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    \[\sin^3x=\sin x \sin^2x\]

  2. anonymous
    • 3 years ago
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    i think the gimmick is to rewrite this as \[\sin^2(x)\sin(x)=(1-\cos^2(x))\sin(x)\]

  3. anonymous
    • 3 years ago
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    then \[\sin(x)-\sin(x)\cos^2(x)\] first integral is easy, second is a u - sub

  4. anonymous
    • 3 years ago
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    or you can look at the "reduction" formula for sine on the inside cover of the back off your text and use \(n=3\)

  5. Trustted
    • 3 years ago
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    HMMMM.. Tnx guys.. can u give me a site for online lessons

  6. UnkleRhaukus
    • 3 years ago
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    \[\sin^3x=\sin x \sin^2x\\\qquad=\sin x\frac{1-\cos(2x)}2\\\qquad=\tfrac12\big(\sin x-\sin x\cos(2x)\big)\\\qquad=\tfrac12\Big(\sin x-\frac{\sin(3x)+\sin(-x)}{2}\Big)\\\qquad=\frac {\sin x}2-\frac{\sin(3x)}{4}+\frac{\sin(x)}{4}\]

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