Is there a better/faster way to take anti-derivatives of simple functions than "reversing" the derivative rules?

Question

Update: For what it's worth, I will wait another few hours to see if anyone else has a more comprehensive answer to my question. But if not, I will "accept" one of the two extant answers, both of which are very good although not quite as comprehensive as I had hoped.

I am taking AP Calculus AB right now and we are learning about anti-derivatives (indefinite integrals) for Unit II. Before that, we learned some basic derivative rules for transformed functions, such as:

$ [f (x + a)]]' = f'(x + a)$

$[f(ax)]'=a*f'(ax)$

(We also learned the derivatives of some elementary functions, e.g. polynomials, exponential functions, sine, and cosine.)

For anti-derivatives we have likewise memorized (and proved) formulas for some basic functions, but unlike with derivatives we have not been taught very much at all about what to do with transformed functions.

Consider, say, finding the anti-derivative $F(x)$ if $f(x)=1/(4x)$. Or finding the anti-derivative $G(x)$ if $g(x)=\cos(4x)$. (Or even worse, how about if $f(x)$ was actually $1/(4x-3)$, and $g(x)$ was actually $\cos(4x-3)$?) I am not entirely sure how to systematically and carefully go about solving such problems.

Should I try to learn integral u-substitution or any tricks like that (which we haven't covered in class yet), or am I better off just trying to intuitively "reverse" the differentiation rules as best I can? I want to figure out a relatively efficient method of integrating basic functions but right now am pretty confused. (Often attempting to reverse the differentiation rules kinda gives me a headache haha and I get utterly lost because it's hard to think about things backwards.)

Answer 1

You will have to learn the substitution rules for more complicated compositions of functions. But for simple things, like $g(x):=\cos(4x-3)$, it is faster to make a reasonable guess, and then "fix the constants". The reasonable guess for the above $g$ is "something with $\sin(4x-3)$". Differentiating $\sin(4x-3)$ gives $4\cos(4x-3)$ with an undesired factor of $4$. It immediately follows that the correct antiderivatives of $g$ are of the form ${1\over4}\sin(4x-3)+C$.

Answer 2

Integrals are really, really hard to the point of some of them that use functions that are simple to understand are impossible to find an antiderivative of (ex. $\int_{}^{}e^{-x^2}dx$ cannot he expressed in any “simple” way. However, there are some more complex methods of integration (such as Cauchy's integral formula) that aren’t necessarily about finding antiderivatives, but those integral formulas are only special cases and you will likely not learn about them over the course of your calculus AB class.

In general, any method of integration you will do in your class will just be applying derivative rules backwards. But over time, in a similar way to how you likely learned to factor expressions quickly in algebra, you will learn to more quickly recognize what to do in a certain integral. But you will still be applying derivative rules backwards. However, if you’re really having trouble I do recommend learning u-substitution and other simpler integral rules. They are essentially just hiding the trial and error under formulas but they do still save a ton of time on more complex integrals.