Hopefully we've gotten used to integration by now and even feel that integrating polynomials is very easy as easy as taking a derivative. This is because integrating a simple polynomial, much like taking a derivative, will follow a simple algorithm that never changes.
But this will not always be the case, as integration doesn't always follow such a rigid algorithm, and in fact there will be times where it is not immediately obvious how to do it.
In such a case there are a few tricks we can use and the one of them we will learn is called the integration by substitution.
Substitution Rule
The integration by substitution allows us to take a complicated integrand and simplify it by representing some portion of it with the letter "u". in this way, integration by substitution is also called as u substitution integration.
But the point is we need to understand thats concept of integration which makes our integral evaluation as easy as we all need. Let's take an overview how to do u substitution ?
How to do U Substitution
Let’s take this problem
$$ \int 2x \; cos(x^2 + 1) \; dx $$
This looks like a mess. The cos is operating on a sum of (x2 +1) and it is also part of a product with "2x".
So nothing we have learned so far will work here. But let's try something cleverly, what if we try an integration by substitution here.
Let's replace as we learn in u substitution
$$ u = x^2+ 1 $$
We are still saying the same thing, Since u = x2+1, but we are simplifying the integrand this way, because if cos is operating on a loan variable "u" rather than an expression "x2+1", we will be able to integrate it.
The only problem is that if we are changing things to "u" then dx will be a problem. We need things in terms of "u". Now we need dx to be du instead.
But we can fix that, let's understand that
$$ \frac{d}{dx}(x^2 + 1) \;=\; 2x $$
Now we can do something tricky
$$ d(x^2 + 1) \;=\; 2x dx $$
We know x2+1 = u, So let's just change that to "u".
$$ du \;=\; 2xdx $$
Now if we bring the 2x inside the integrand to the other side of the cos term.
we have 2x dx on right side, here
$$ \int cosu \; 2xdx $$
So according to these equations,
u = x2+1 and du = 2xdx
we can just rewrite that
$$ \int cosu \; du $$
Now we are ready to integrate this
So, the integral of cos u will be
$$ sinu + c $$
Now that integration is completed by using u substitution integration. Let's return you back to its original form u substitution to orignal complex function and we have
$$ sin(x^2 + 1) + c $$
That’s the final answer.
Relation B/w Chain Rule and Integration by Substitution Rule:
Just to be thorough let's differentiate this and make sure we get what we started with
$$ \left[ sin(x^2 + 1) + c \right]' $$
Since C is a constant, it will become zero. So By using the Chain Rule of Differentiation.
$$ \left[ sin(x^2 + 1) \right]' = cos(x^2 + 1)(x^2 + 1)'$$ $$ \left[ sin(x^2 + 1) \right]' = cos(x^2 + 1)(2x)$$
That is indeed what we started with. So, u substitution integration worked like a charm.
In this way we can think of the "u substitution rule as the chain rule in Reverse" because we symbolize some piece of the function with a single variable allowing us to integrate more simply, just the way that the chain rule ignores the inner function during the first step of differentiation.
The only thing to remember is that while the chain rule will always work but the u substitution rule won't always work.
integration by substitution only worked where our expression for du was indeed present in the integrand because the integrand was in the form of f(g(x)) times to g'(x).
$$ \int f[g(x)] \; g'(x) dx $$
The integrand must be in this format for work on this method.
U Substitution Examples:
Let’s Try Another Example:
$$ \int x^2 \; \sqrt{x^3+1} \; dx $$
Is this integrand in the form that we mentioned ?
$$ \int f[g(x)] \; g'(x) dx $$
If we suppose, f = square root and g(x) = x3+1, then g’(x) will be 3x2. We Know we have x2 here. So, Let’s Try and see the solution.
Let's substitute inner function as we discussed in u substitution integration:
$$ u = x^3 + 1 $$ $$ \frac{du}{dx} = 3x^2 $$
That means,
$$ du = 3x^2 dx $$
Now, we have only x2 dx in integrand, that’s not an issue we simple take 3 to the other side.
$$ \frac{du}{3} = x^2 dx $$
Now by using "u" and "du", our integral will be :
$$ \int \frac{1}{3} \sqrt{u} \; du $$
Now Let’s integrate our easiest form of integrand. We can write it as
$$ \frac{1}{3} \int u^{\frac{1}{2}} \; du $$
This will become,
$$ \implies \frac{1}{3} \times \frac{2}{3} \; u^{\frac{3}{2}} $$
Now Simple Return back the value of u, So the final answer by using substitution rule is:
$$ \frac{2}{9}(x^3+1)^{\frac{3}{2}} \;+\; c $$
That is done now.
Final Words
So remember anytime we are integrating some function, and the function can be split up into a smaller function followed by the derivative of that smaller function, or something very close to it, the substitution rule is a good idea for something to try.
This won't always work, but if it does, it can evaluate the complicated integral and make it very simple (As we see in u substitution examples), so it is definitely a great start to our bag of tricks for integration.