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About LinkBacksSo im still having trouble integrating functions like this:
:integration: (x(sin5x))dx
Am i supposed to use some sort of reverse-product rule? i tried substitution,the only integration tech i know yet.
Integration by Parts is the most direct method, but if you don't know it, you can always look at the relationship between this function and its derivative.Originally Posted by Nervous
and
So working in reverse, that means we have
![\displaystyle \begin{align*} \int{\left[ \cos{(5x)} - 5x\sin{(5x)} \right] dx} &= x\cos{(5x)} + c \\ \int{\cos{(5x)}\,dx} - 5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} + c \\ \frac{1}{5}\sin{(5x)} - 5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} + c \\ -5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} - \frac{1}{5}\sin{(5x)} + c \\ \int{x\sin{(5x)}\,dx} &= \frac{1}{25}\sin{(5x)} - \frac{1}{5}\,x\cos{(5x)} + C \textrm{ where } C = -\frac{1}{5}\, c \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \int{\left[ \cos{(5x)} - 5x\sin{(5x)} \right] dx} &= x\cos{(5x)} + c \\ \int{\cos{(5x)}\,dx} - 5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} + c \\ \frac{1}{5}\sin{(5x)} - 5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} + c \\ -5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} - \frac{1}{5}\sin{(5x)} + c \\ \int{x\sin{(5x)}\,dx} &= \frac{1}{25}\sin{(5x)} - \frac{1}{5}\,x\cos{(5x)} + C \textrm{ where } C = -\frac{1}{5}\, c \end{align*})
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