Integration By Parts (Level 3) Alright, we are going to continue covering
integration by parts, this time around we will go over problems that require multiple
applications of the integration by parts formula and those that generate a constant multiple
of the integral, let's try the former type of problems first:
Find the integral of x squared times cosine of x dx. Ok, so this time around we actually have
a product of two functions x squared and cosine of x. Like always we need to assign our u
and dv. The best choice occurs when we let u equal to x squared because this function becomes
simpler when we take its derivative so taking the derivative of u with respect to x we obtain
2x which is 1 power less than the original function and as a result a much simpler function.
Next we assign dv equal to cosine of x dx taking the integral of cosine of x we obtain
sine of x. Now that we have all the required functions we use the integration by parts
formula and substitute them, doing that we have that the integral of x squared times
cosine of x is going to be equal to:
x squared times sine of x minus the integral of sine of x times 2x dx.
now notice that the integral of sin of x times 2x requires an additional application of the
integration by parts formula, what we are going to do is to carry out a substitution
and let the variable I equal this integral that requires an additional application of
the integration by parts formula, it's important not to lose track of the original integral
we are looking for, so making the substitution we have the following expression. Now putting
that aside and focusing on this integral we carry out the familiar steps and apply the
integration by parts formula once more. Same ideas apply make u equal to a function that
becomes simpler when differentiated so we let it equal to x, in the same manner we let
dv equal to sine of x so taking the integral of sin of x gives us negative cosine of x, substituting
these expressions into the integration by parts formula we have the following:
the integral of x times sine of x dx equals, negative x times cosine of x minus the integral of negative cosine of x dx.
we factor out the negative
sign and simplify, and finally taking the integral of cosine of x we have that the integral
of x times sine of x dx is equal to negative x times cosine of x plus sine of x plus C.
Now lets go back to the original integral that we are trying to solve and substitute
I with this result that we just obtained and then we simplify doing that we obtain the
following, notice that we ended up with twice the constant here, keep in mind that
if you take a constant and double it you still obtain a constant so we can replace 2C with
another constant. So after all that it turns out that the final integral of
x squared times cosine of x dx
is equal to
x squared times sine of x plus 2x times cosine of x minus 2 times sine of x plus C one.
where C one is just equal to twice the original constant
a word of caution, when making repeated applications of the integration by parts, you need to be
careful not to interchange the substitutions in successive applications for example in
the this problem we decided to let u equal to x and dv equal to sin of x, if in the second
application we decided to switch the substitutions with u being equal to sin of x and dv equal
to x we would have obtain the following:
the integral of x times sine of x dx will be equal to:
sine of x times one half x squared minus the integral of one half x squared times cosine of x dx.
and I will be represented by:
x squared times sine of x over 2 minus the integral of x squared cosine of x over 2 dx.
and substituting this expression to the original
integral we would obtain:
x squared times sine of x minus x squared times sine of x plus the integral of x squared times cosine of x dx.
at this point notice that by switching the second substitutions
we undue the previous integration and returned to the original integral. So be careful when
you repeatedly apply the integration by parts formula.
Repeated use of the integration by parts formula seems pretty tedious at times, there is
actually a little short cut that can save you some time,
it is often referred to as the Tabular method, this method is best
suited for integrals of the form:
x to the n times sine of a times x dx, or
x to the n times cosine of a times x dx, or
x to the n times e to the power of a times x dx, or
Let's use the tabular method to find the integral of the following problem:
Find the integral of x squared times sine of pi times x dx.
First step is to set up a table with 2 columns, the first column will contain u and its derivatives
and the second column will contain dv and its integrals or antiderivative.
Next we select our u and dv for this integral we set u equal to x squared, now the next
step is repeatedly take derivatives until we obtain 0 doing that yields the following
functions next we take our dv and repetadtly take integrals to match the number of rows
that were generated from the differentiation step in this case we take 4 integrals one
after the other doing that yields the following functions. Next, it's just a matter of pairing
up the functions, so we take the function from column 1 and multiply it diagonally to
the right by the function on the second column and 2nd row, next we multiply the function
in the 1st column and 2nd role with the function in the 2nd column and 3rd role, and we repeat
this pattern until we reach the function in the 2nd column and last row. Next we alternate
the signs going down the middle between both columns for each row as follows, +, - , +
and include them when we multiply the diagonal entries, so in the first product would be
a + times + times a negative which results in a negative expression, in the second product
we have a + times a negative times a negative which results in a + expression, and finally the
last product we have a +, times a + times a + which results in a + expression. Once
all of this has been taking care of, we essentially have our answer.
The integral of x squared times sin of pi times x dx will be equal to:
negative x squared times one over pi times cosine of pi times x plus
2x times one over pi squared times sine of pi times x
plus 2 times one over pi cubed times cosine of pi times x plus C
Alright, in our next video we will go over
an example that generates a constant multiple of the original integral and use integration
by parts formula along with the tabular method to find it's integral.