u substitution and du in integrals?
Olivia Zamora 
These were a couple examples in class before learning the $u$-substitution method for integrals. I'm not sure what is going on:
$$\int1\, d(2x) = 2x+ C$$ $$\int \sin x \,d(\sin x) = \frac{(\sin x)^2}{2}+C$$ $$\int(2x+1) \,d(2x+1) = \frac{(2x+1)^2}{2}+C$$
How exactly do we reach these answers? I guess I can see how the last two follow if you substitute in $U$ and use the power rule, but what about the first one?
I guess the $d(\dots)$ is throwing me off. Of it's $\int 1 \,dx$, the answer would be $x+c$, but with $d(2x)$, the answer is $2x+C$. What is happening? How would you find the integral $\int 3 \,d(2x)$?
$\endgroup$2 Answers
$\begingroup$This is actually a fairly advanced topic called Riemann-Stieltjes integrals. You've been given some pretty easy examples, however.
When you have an integral $\int f(x)\, dx$, you're saying, "the integral of $f(x)$ with respect to $x$." The "with respect to" is the key part. It suggests that we could take these integrals with respect to something else...
In $\int 1\, d(2x)$, we're taking the integral of $1$ with respect to $2x$. You can think of this as having $2x$ as our base function, rather than $x$. To evaluate these simply, do what you suggested and do a substitution. Let $u=2x$, then
$$\int 1\, d(2x)=\int 1\, du=u+C=2x+C.$$ Letting $u=\sin(x)$, $$\int \sin(x)\, d(\sin(x))=\int u\, du=\frac{u^2}{2}+C=\frac{\sin^2(x)}{2}+C.$$ And finally, letting $u=2x+1$, $$\int 2x+1\, d(2x+1)=\int u\, du=\frac{u^2}{2}+C=\frac{(2x+1)^2}{2}+C.$$
More on these integrals if you're interested:
$\endgroup$ $\begingroup$I had the same question, also superklc's answer is a good one. Let me just add on:
Note:
$$\frac{\ d (f(x))}{\ dx}=f'(x)$$
And consequently notationally we can say:
$$\ d(f(x))=f'(x) \ dx$$
So whenever you see $\ d (f(x))$ it might help you to replace it by $f'(x) \ dx$.
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