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Brooke_army

  • 3 years ago

Integration by parts

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  1. Brooke_army
    • 3 years ago
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    \[\int\limits_{}^{} \cos(-10x)dx\] u=-10x

  2. Mimi_x3
    • 3 years ago
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    you dont need IBP

  3. Brooke_army
    • 3 years ago
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    what?

  4. Mimi_x3
    • 3 years ago
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    \[\int\limits cosaxdx =\frac{1}{a} sinax+c\]

  5. RadEn
    • 3 years ago
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    just use the basic formula : int (cosAx) dx = 1/A * sinAx + c

  6. Brooke_army
    • 3 years ago
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    I don't understand the basic formula

  7. anonymous
    • 3 years ago
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    @Brooke_army "parts" is something else that you may not have seen yet. this type of integration is usually called "u - substitution"

  8. anonymous
    • 3 years ago
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    if you didn't have the \(-10x\) you would have \[\int cos(x)dx=\sin(x)\] because \[\frac{d}{dx}[\sin(x)]=\cos(x)\] but \[\frac{d}{dx}[\sin(-10x)]=-10\cos(-10x)\] by the chain rule therefore, to get what you want, you have to divide by \(-10\)

  9. anonymous
    • 3 years ago
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    so if you take the derivative of \(-\frac{1}{10}\sin(-10x)\) you get exactly what you want, namely \(\cos(-10x)\)

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