anonymous
  • anonymous
Integration by parts
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{}^{} \cos(-10x)dx\] u=-10x
Mimi_x3
  • Mimi_x3
you dont need IBP
anonymous
  • anonymous
what?

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Mimi_x3
  • Mimi_x3
\[\int\limits cosaxdx =\frac{1}{a} sinax+c\]
RadEn
  • RadEn
just use the basic formula : int (cosAx) dx = 1/A * sinAx + c
anonymous
  • anonymous
I don't understand the basic formula
anonymous
  • anonymous
@Brooke_army "parts" is something else that you may not have seen yet. this type of integration is usually called "u - substitution"
anonymous
  • anonymous
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 \(-10\)
anonymous
  • anonymous
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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