Can someone explain this please? If $\displaystyle \int$ udv = uv - $\displaystyle \int$ vdu then how does I just don't see why the last part the 48cos(t/4), did not stay as the sine.
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Originally Posted by gammaman Can someone explain this please? If $\displaystyle \int$ udv = uv - $\displaystyle \int$ vdu then how does I just don't see why the last part the 48cos(t/4), did not stay as the sine. Because $\displaystyle \int \sin \frac t4 ~dt=-4 \cos \frac t4+c$
Oh I see, because of the $\displaystyle \int$ after the minus sign, I must still integrate and change the sine to cosine, right?
Originally Posted by gammaman Oh I see, because of the $\displaystyle \int$ after the minus sign, I must still integrate and change the sine to cosine, right? Yes (I made a small typo, the integral of sine is -cosine)
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