To evaluate more complex integrals we need to learn a few tricks most of these are rules of integration that correspond to particular rules of differentiation the substitution rule which we just learned works kind of like the chain rule for differentiation the product rule for differentiation also has a corresponding technique for integration and it is called integration by parts.
Recall the Product Rule of Differentiation:
So let's learn this technique now. Let's quickly recall the product rule, it says that the derivative of a product of two functions is equal to the first times the derivative of the second plus the derivative of the first times the second. Mathematically,
$$ \left[ f(x)g(x) \right]' \;=\; f(x)g'(x) \;+\; f'(x)g(x) $$
Well we know that we can always go backwards by integrating. So the integral of the right hand side derivative must be the product of the two functions that we started with.
$$ \int f(x)g'(x) \;+\; f'(x)g(x) \;=\; f(x)g(x) $$
Now let's work with this expression a little bit to get something more useful.
Integration By Parts:
We know that the integral of a sum is equal to the sum of integrals. So let's split this integral up into two separate ones
$$ \int f(x)g'(x) \;+\; \int f'(x)g(x) \;=\; f(x)g(x) $$
Then let's bring the second integral over to the other side of the equation by subtraction and there we have integration by parts formula.
$$ \int f(x)g'(x) \;=\; f(x)g(x) \;-\; \int f'(x)g(x) $$
It may look like we haven't done much but this is actually quite powerful, because up until now, we haven't had any way to integrate a product of two functions other than the substitution rule.
We can see that if we have the product of two functions to integrate we simply list the first function times the antiderivative of the second function and subtract from that the integral of the antiderivative of the second function times the derivative of the first function.
An easier way to remember all of this is to represent if:
f(x) = u and g(x) = v , then
f'(x)dx = du and g'(x)dx = dv
So, the Integration by parts will be written mathematically as
$$ \int u \; dv \;=\; uv - \int v \; du $$
The thing to keep in mind here is that while we can apply the technique to any integrand that is a product of functions, sometimes this will make things simpler and sometimes it won't.
You'll notice that the technique leaves us with a different integral to evaluate and if this one is no simpler than what we started with then this technique is not very useful.
But let's look at some integration by parts problems where this approach works like a charm and you'll get a better sense of exactly when to use integration by parts.
Integration By Parts Examples
So to learn that how to do integration by parts, we should need to evaluate some integration by parts problems, which will be helpful in the sharp understanding of our concept.
- Let’s try to integrate
- $$ \int x \; sinx dx $$
Before we start getting used to u’s and v’s. Let's just use the first integration by parts formula with f(x) and g(x).
Here's a hint: we know that between these two functions x and sinx one of them will be f(x) and the other will be g'(x).
The trick is that we want the function that becomes much simpler when differentiated to be f(x) because f’(x) will remain in the integral.
$$ \int f(x)g'(x) \;=\; f(x)g(x) \;-\; \int f'(x)g(x) $$and we will want to be able to evaluate this integral easily. So let's use f(x) = x because its derivative is 1 which is as simple as it gets.
So in the first integral,
f(x) = x and g'(x) = sinx
By Using f(x) and g'(x)
f'(x) = 1 and g(x) = (-cos x)
Now putting all these values in integration by parts formula for evaluating
Since,
$$ \int f(x)g'(x) \;=\; f(x)g(x) \;-\; \int f'(x)g(x) $$By putting values
$$ \int x \; sinx \; dx \;=\; -x \; cosx \;-\; \int (- cosx)(1) \; dx $$As we just said, here's the best part f’(x) = 1. So the second integral is simply negative cosx. We can also take out the negative sign to cancel out the first one.
$$ \int x \; sinx \; dx \;=\; -x \; cosx \;+\; \int (cosx) \; dx $$As cosx is trivial to integrate its sinx
$$ \int x \; sinx \; dx \;=\; -x \; cosx \;+\; sinx \; $$so we get final answer here
$$ \int x \; sinx \; dx \;=\; -x \; cosx \;+\; sinx \;+\; c $$So bear in mind that the reason this worked is because we chose the function for f(x) that would yield f’(x) that would make this second integral much easier to evaluate than the first.
- Let’s try another integration by parts example
- $$ \int \; lnx \; dx $$
Now we just use the integration by parts formula in terms of u's and v's
$$ \int u \; dv \;=\; uv - \int v \; du $$As you can see we have only one function to integrate but we can solve it by using integration by parts. Because ln x is multiplied by simple 1.
We can write
u = ln(x) and dv = 1 dx
From u and dv,
du = (1/x)dx and v = x
Now putting back these values in the formula of integration by parts
$$ \int \; lnx \; dx \;=\; x \; lnx \;-\; \int \; x \frac{1}{x} \; dx $$So, x is cancelled by 1/x in the second integral
$$ \int \; lnx \; dx \;=\; x \; lnx \;-\; \int dx $$Since the integral of dx is x. So the final answer will be:
$$ \int \; lnx \; dx \;=\; x \; lnx \;-\; x \;+\; c $$
So these are some integration by parts examples which help use to understand when to use integration by parts.
Conclusion:
So integration by parts is just another technique in our bag of tricks that we can use to evaluate integrals.
The substitution rule works well when the integrand is the product of a function and its derivative or close to it. But the integration by parts works when the two functions are not related at all but one of them becomes much simpler when differentiated, as we see in integration by parts problems, and it makes the new integral easier to solve than the original one.