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

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.

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.

• $3dx = d(3x-1)$, I believe this is missing here. Commented Sep 4 at 14:39
• So we are just doing the following: $$\int\cos(3x-1)dx=\frac{1}{3}\int\cos(3x-1)\cdot 3dx=\frac{1}{3}\int\cos(\phi)\frac{d\phi}{dx}dx=\frac{1}{3}\int\cos(\phi)d\phi=\frac{1}{3}\sin(\phi)$$ Am I understanding this correctly? Commented Sep 4 at 14:44
• Yes, it is that. Commented Sep 4 at 14:57

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.