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How does the integral sin5x = -1/5cos5x. I just do not understand where the 1/5 comes from x.x

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\[\int\limits_{b}^{a} \cos 5x= \frac{ -1 }{ 5 }\sin 5x (a \to b)\]
just treat 5x as one thing and then use property of integration which says coefficient gets denominator so 5 comes underneath sin 5x
Yeah I do know its like a chain rule ( so treat sin5x than 5x.). However when you have a coefficient (5x) i thought it would go as 5x = 5x ^2/2

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nope it wont go that way
Use a substitution: \[u=5x\Rightarrow du=5\;dx\Rightarrow \frac{1}{5}du=dx\] In the integral, you have \[\int\sin(5x)\;dx=\int\sin(u)\;\left(\frac{1}{5}\;du\right)=\frac{1}{5}\int\sin(u)\;du\] which is equal to \[-\frac{1}{5}\cos(u)+C=-\frac{1}{5}\cos(5x)+C\]
In short, the 1/5 comes from the substitution for dx, which depends on the sub you make for 5x.
ahhh the substitution rule makes sense x.x stupid me. THanks!

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