Using the definition of the substitution rule to calculate the antiderivative of a function

Question

My task is the following:

Find all antiderivatives of $\cos(3x-1)$.

I instantly thought that I have to do that with the substitution rule, but I don't understand the definition that is given and I am not able to use it to come to a solution:

Let $f:I\rightarrow\mathbb{R}$ be continous and $\phi:[a,b]\rightarrow I$ continous differentiable. Then the following holds: $$\int_a^b f(\phi(t))\cdot\phi'(t)dt=\int_{\phi(a)}^{\phi(b)}f(x)dx$$

I now tried to solve my task with exactly this definition. For this I defined two functions $f$ and $\phi$ like the following: $$f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\cos(x)$$ $$\phi:\mathbb{R}\rightarrow\mathbb{R}, \phi(x)=3x-1$$

Now already my first confusion started, because if I am strictly following the definition I thought that I wouldn't be allowed to use substitution here, because $\phi$ has the real numbers as domain which is not a closed interval. For the first I tried to ignore this, but I would be happy if someone could explain here if I am still allowed to use substitution or not.

I then of course checked if everything else holds and stated that $f$ is continous and $\phi$ is continous differentiable (if I am correct that means that the differentiate of $\phi$ is continous).

Now I tried to calculate $\int\cos(3x-1)dx$ and here of course my second confusion started because I have no bounds on the integral and thus it doesn't match exactly the given definition. I also ignored this for the beginning but I would be happy if someone could explain if I am still allowed to use the definition here.

Now I finally came to rewriting the integral where I tried to strictly follow the definition. And here I just didn't came to a way that would make sense.

First I know that I have to calculate the integral $\int\cos(3x-1)$ to get to the antiderivative of $F(x)+c$ where $c\in\mathbb{R}$. So I tried to rewrite this integral such that I could be able to use the substitution rule:

$$\int\cos(3x-1)\overset{?}{=}\frac{1}{3}\int\cos(3t-1)\cdot3 dt=\frac{1}{3}\int\cos(x)dx=\frac{1}{3}\sin(x)+c$$

This is obviously the wrong result, but I don't understand how I can come to the correct solution $F(x)=\frac{1}{3}\sin(3x-1)+c$. The question mark about the first equals sign should indicate that I am not sure if this equality really holds (I think it only does for $x=t$). I looked at other sites and I know that I have to set $t=3x-1$ but I don't see how I would come to such a conclusion by strictly following the me given tools for this task. Which is only the description of the task exactly like I have cited here and the definition of the substitution rule also exactly like I cited here.

You can help me with explaining how I can come to the right result by following the me given tools and also by talking a bit about my two confusions. Also I am not given any examples by my teacher, so maybe you can try to explain my difficulties by another example at first so I can try to solve the problem from the example.

Answer 1

You did the integral almost right. Just extra caution is needed. Let us use a definite integral to avoid confusion. In the end, we will just drop the limits of integration. $$ I = \int_a^b \cos(3x-1)dx=\frac{1}{3} \int_a^b \cos(3x-1)\cdot 3 dx $$ where we used $\phi(x) = (3x - 1)$. Then, as the definition says: $$ I = \frac{1}{3}\int_{3a-1}^{3b-1} \cos(\phi)d\phi = \frac{1}{3}\sin(\phi) \Big|_{3a-1}^{3b-1} $$ Dropping the limits of integration and substituting back $(3x-1)$ for $\phi$ we arrive at the desired result $F(x) = \frac{1}{3}\sin(3x-1)$.

As for an explanation of definition, you can understand it as follows: $\phi' = \frac{d\phi}{dt}$, so we can (symbolically) write $\phi' dt = \frac{d\phi}{dt} dt = d\phi$, cancelling $dt$'s. Limits of integration just change with the change of integration variable.

Answer 2

You want to find the primitives of the function $x \mapsto \cos(3x-1)$. The goal here is to make appear a function that you know you can integrate.

Let $u=3x-1 \Longrightarrow du=3~dx \Longleftrightarrow \frac{du}{3}=dx$

It comes that,

$$ \int \cos (3x-1)~dx = \frac{1}{3}\int \cos(u)~du = \frac{1}{3} \sin (u) + c = \frac{1}{3} \sin (3x-1) + c$$ where $c \in \mathbb{R}$ is a constant. The substitution here is not mandatory since $x \mapsto 3x-1$ is linear, meaning that its derivative is a constant. Just multiply the integral by $3$ and $1/3$ and you can immediately integrate.