in this video we're going to compute this integral of sine x cubed and we don't have any
way of getting the differential of x cubed tacked onto this thing so we're just basically stuck
unless we use an infinite series trick in this case so we're going to start with a reminder of
the standard power series for the sine of x then we're going to modify that to turn it into the
sine of x cubed and then we're going to integrate the power series term by term and we'll get an
infinite series expansion as our solution for this otherwise impossible integral so here's a reminder
of the power series for the sine function i'm going to write this first in expanded form it's
all the odd powers of x each of them comes with that same number factorial and the signs alternate
but it's going to be really useful to see this in sigma notation so i'm going to write it that way
too i want to start with a positive term so i have negative 1 to the n minus 1 so that when i plug in
n equals 1 i start with negative 1 to the 0 which is 1. and then to get these odd powers of x so i'm
starting with a power of 1 and then 3 and then 5 and so on and so on i can use 2n minus 1 so when
i sub in n equals 1 that exponent gives me a one sub in x equals two it gives me a three and so on
and then the factorial in the denominator always matches that power so it's the same expression
so there's our classic power series for the sine function and i'll post a link
real quick to the video where i first derived this and now we're going to modify that series
into the sine of x cubed so we just replace all the x's with x to the third and i end up with x
cubed minus x cubed cubed and a power to a power you multiply the exponents so i have
x to the ninth over three factorial plus x cubed to the fifth that's x to the fifteenth
over five factorial minus x to the third to the seventh that's x to the 21st over seven factorial
and so on and so on and then in the sigma notation i'm just replacing x with x cubed
so i have x cubed to the 2n minus 1 and you multiply those exponents so that gives me x
to the six n minus three now that i'm thinking of the integrand here as an infinite series
i can just integrate that infinite series term by term and so if i look at that in
expanded form first i had x cubed and that integrates to x to the fourth over four
and i had x to the ninth over three factorial and that integrates to x to the tenth over ten times
three factorial and i had x to the 15 over five factorial which integrates to x to the 16 over
16 times 5 factorial but i can also just think of integrating term by term inside the sigma notation
and if i have an x to the 6 n minus 3 i just add 1 to that exponent and i get x to the 6 n minus 2
and then i divide by that exponent this is just the power rule for antiderivatives and i've got
immediately in sigma notation an infinite series representation of the value of this antiderivative
i should also include that there is an arbitrary constant of integration here
and then the last thing i wanted to do is just verify that this sigma notation
is working that when i plug in n equals 1 2 and 3 i actually get these terms 1 2 and 3
for the expanded form so plugging in n equals 1 i get negative 1 to the 0 so i get a positive term
n equals 2 gives me a negative term and so on so i'm happy with my signs
and when i plug in that n equals 1 into my x term i end up with 6 minus 2 for the exponent or
4. so i have that working and in the denominator i get a 6 minus 2 here which is 4 times 2 n minus 1
or 1 factorial so the denominator comes out to 4 so the first term works if i just look
at one more of these plugging in n equals 2 in my x term i get x to the 12 minus 2 or x to the
10th which i see here so that's good and then the first term in the denominator is going to be 10
and i see that right here and the second term in the denominator plugging in n equals 2 gives me 3
factorial so that works as well so it looks like everything is working out so this is one of many
cases in applied mathematics where we're making a trade-off and the trade-off we've made here is
we ran into an integral that was just simply impossible to do with the analytical methods
we had prior to this so we decided to expand this thing in an infinite series and integrate
term by term and we sort of got the job done but the cost of this is that we now have an infinite
series representation of our final function instead of just a nice analytical formula for it
but certainly if this was a definite integral we could use this infinite series and keep as many
terms as we needed to get an answer as precise as we wanted it's just not going to be exact
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