Integration By Parts (Level 1)
Integration is not as straightforward as differentiation; there are no rules that absolutely guarantee
obtaining an indefinite integral of a function. In the following videos we will learn various
integration techniques to help us find indefinite integrals of more complicated functions.
Back in calculus I we learned our first integration technique specifically the substitution rule.
Recall that every differentiation rule has a corresponding integration rule. For example,
the substitution rule for integration corresponds to the chain rule for differentiation.
Likewise the rule that corresponds to the product rule for differentiation is called
the rule for integration by parts
If we take the product rule and rewrite it using the notation for indefinite integrals
we would have the following expression, remember according to the fundamental theorem of calculus,
differentiation and integration are inverse processes so we can rewrite the product rule
as follows, if we were to take the integral of the resulting expression of the product
rule we would obtain back the original function formed by the product f of x and g of x.
Again because the fundamental theorem of calculus tells us that differentiation and integration
are inverse processes so this expression is the product rule written by using indefinite
integral notation. Now here what we do is distribute the integral and dx operator and
we end up with the following expression. Finally we solve for the integral of one of the products,
in this case I will solve for the expression all the way on the left hand side, so rearranging
the expression results in following formula.
The integral of f of x times g prime of x dx is equal to
f of x times g of x minus the integral of g of x times f prime of x dx
This formula is called the formula for integration
by parts, and you can easily derive it by starting with the product rule and taking
the integral of both sides and rearranging the equation. So there is no need to memorize
this equation. An alternative way of writing this formula
is by carrying out the following substitutions if we let u equal f of x and v equal g of x then the differentials of the
functions are du equals f prime of x dx
and dv equal g prime of x dx respectively.
Substituting these expressions we have an alternative form
of the integration by parts formula specifically
The integral of u dv is equal to u v minus the integral v du
Alright now that you know the integration by parts formula when and how do you use it?
First of all when do you use it, usually when you encounter an integral of a product of
functions such as the integral
x times cosine of x dx
Notice that we have a product of two functions specifically x and cosine of x. This is a
perfect candidate for integration by parts. Alright now Let's find the integral by using
integration by parts formula we will first solve it by using the first version of the
formula
In order to use and apply this formula to our problem we need make a decision and determine
which of the two functions will f(x) equal to and which one will g prime of x equal to.
This an important decision, the main goal in using integration by parts is to obtain
a simpler integral than the one we started with. Let's go over the correct decision,
the simplest integral is obtain If we let f of x equal to x and let g prime of x equal
to cos of x. By making this decision it turns out that f prime of x after taking a derivative
of f of x will be equal to one and g of x is equal sin of x after taking the integral of
g prime of x remember taking the integral of a derivative leads you to the original
function g of x.
Now we substitute each expression into the integration by parts formula so we have that,
f of x times g of x becomes x times sin of x and the integral of g of x times f prime
of x becomes the integral of sin of x times one. Here the integral of Sin of x is just equal
to negative cosine of x plus the constant remember we are working with an indefinite integral
and we need to include the constant. So the final answer is x times sin of x + cosine
of x + C
Lets go back to the first step and see what happens when make the wrong decision when
determining our f of x and g prime of x. If we let f of x be equal to cos of x and g prime
of x be equal to x then our f prime x will be equal negative sin of x after taking the
derivative of f of x, and our g of x will be equal to x squared over two after taking
the integral of g prime of x. Now the problem arises when we substitute these expressions
into the integration by parts formula. Doing that we have the following expression,
The integral of x times cosine of x dx is going to be equal to cosine of x
times x squared over two
minus the integral of negative x squared over two times sine of x dx
Notice that we actually ended up with a slightly more complicated integral than the one we
started with this is a tell tell sign that you assigned your f of x and g prime of x
functions wrong the main point is to obtain a simpler integral. At this point you should
start over and switch the corresponding functions. A tip to avoid starting over is by mentally
taking the derivative of your assigned f of x function and taking the integral of your
assigned g prime of x function and check to see if those functions become simpler in this
case the function becomes simpler if we take the derivative of x as oppose to taking the
integral which yielded a higher degree power function in this case x squared over two it's
much easier to work with the function one than the function x squared over two.
Before we end the video lets solve this same integral using the alternative form of the
integration by parts formula, generally this is the form that most college classes teach
in my opinion using the first form of the integration by parts formula is the way to
go. Nevertheless some students like this second form because it is easier to remember for
those students that like to memorize there way trough math classes which I personally
don't recommend. OK. The first step in using the second form
of the integration by the parts formula is to make a decision and assign a function for
u and a function for dv so we let u = x and we let dv = cos x dx notice that when we assign dv
we are also assigning the differential dx so keep that in mind and make sure you include
it in your substitution. Now we take the derivative of u with respect to x and solve for du by
multiplying both sides of the expression by the differential dx. So we have that du = dx,
next we take the integral of dv so we obtain v which is equal to sin x now we substitute
u v, du and dv into second form of the integration by parts formula. u v becomes x times sin
of x and v du becomes sin of x times dx. This is the same expression that we obtain
by using the first formula so taking the integral of sin of x and simplifying we obtain a final
answer of x times sin of x + cosine of x + C its really up too which form you want to
use personally the first formula is a lot easier to work with but the second might be
easier to remember. Ok in our next video we will go over more examples involving integration
by parts.