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OpenStudy (anonymous):
\[\frac{ \sin(6x) }{ 6 } + C\]
because its derivative is cos(6x)
zepdrix (zepdrix):
:O
OpenStudy (anonymous):
how did u get that?? @Euler271
OpenStudy (anonymous):
???? @zepdrix
OpenStudy (anonymous):
what's the derivative of cos(6x)?
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OpenStudy (anonymous):
-sin6x ???? @Euler271
OpenStudy (anonymous):
-6sin6x
OpenStudy (anonymous):
the integral is the inverse operation of the derivative.
so the integral of -6sin6x must be cos6x.
tbh i dont even know how to explain it lol. i hope this helps
OpenStudy (anonymous):
yea sorry!! I get u! thanks!@Euler271
OpenStudy (anonymous):
^_^
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zepdrix (zepdrix):
That's a good method to get comfortable with, just sort of like.. advanced guessing.
You know that the derivative will produce an extra factor of 6 because of the 6x inside of the trig function.
So when we integrate, we need to divide by a factor of 6.
If this is still confusing though, you can certainly perform a u-substitution:\[\Large \int\limits \cos6x\;dx,\qquad\qquad u=6x\]