# Difference between the two substitution methods in integration

There exist two methods of integration by substitution.

The first is: While the second is: Unfortunately, I'm having some trouble in understanding clearly the difference between the two; they appear to me as the same thing. Can someone explain to me (possibly by words, I'm not asking mathematical rigorous demonstrations) where is intuitively the actual difference?

## 2 Answers

One way of looking at these is that Theorem 1.18 discusses "straightforward substitution", while an example of Theorem 1.27 is trigonometric substitution.

For Theorem 1.18, consider for example $$f(u) = \frac{1}{2}\cos u$$, which has antiderivative $$F(u) = \frac{1}{2}\sin u$$. Let $$\varphi(x)=x^2$$. Then to integrate $$x\cos x^2$$, we get $$\int x\cos x^2\,dx = \int f(\varphi(x))\varphi'(x)\,dx = F(\varphi(x)) = \frac{1}{2}\sin x^2,$$ which is correct.

For Theorem 1.27, consider instead trigonometric substitution. For example, let $$f(x) = \frac{1}{\sqrt{1-x^2}}$$, and let $$\varphi(t) = \sin t$$. Then $$\int f(x)\,dx = \int f(\varphi(t))\varphi'(t)\,dt = \int\frac{\cos t}{\cos t}\,dt = t+C\,dt.$$ Substituting $$\varphi^{-1}(t) = \arcsin x$$ for $$t$$ gives $$\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C,$$ which is correct.

• Thanks for the answer! – Shootforthemoon Dec 7 '19 at 16:42
• YW. If it fits your needs, you can accept the answer by clicking the check mark below the voting toggles. – rogerl Dec 7 '19 at 16:43

The differing usage of the symbols in the two theorems may cause some confusion.

I will stick to the same symbols across the two theorems.

Let's start with the function $$f(x)$$ and it's antiderivative $$F(x) = \int f(x)\; dx (+C)$$.

Besides this, let's write the $$\varphi$$ from the theorem in the more convenient form $$x=x(t)$$.

Case 1:

Theorem 1.18 refers to the case that you know $$F(x)$$. Then, you also know the antiderivative of $$f(x(t))\frac{dx}{dt}$$

$$\int f(x(t))\frac{dx}{dt}\; dt = F(x(t)) (+C)$$

This is just a direct consequence of the chain rule of differentiation.

Case 2:

Theorem 1.27 refers to the case that you don't know $$F(x)$$, but you know the antiderivative

$$\int f(x(t))\frac{dx}{dt}\; dt = G(t) (+C)$$

Now, the question is, how does $$G(t)$$ relate to $$F(x)$$?

If you can invert $$x=x(t)$$, then you you can write $$t= t(x)$$ (this plays the role of $$\varphi^{-1}$$ from the theorem). Now, having in mind that $$x(t(x)) = x$$ and $$\frac{dx}{dt}\cdot \frac{dt}{dx} = 1$$, the chain rule gives

$$\frac{dG(t(x))}{dx} = \frac{dG(t(x))}{dt}\frac{dt}{dx} = f(x)\frac{dx}{dt}\cdot \frac{dt}{dx}=f(x)$$

This means, if you can invert $$x=x(t)$$, the searched for antiderivative of $$f(x)$$ is $$F(x) = G(t(x))$$.