### Video Transcript

Use the Maclaurin series of sin π₯ to express the integral of the sin of π₯ cubed with respect to π₯ as an infinite series.

The question wants us to use the Maclaurin series of the sine function to express the integral of sin π₯ cubed with respect to π₯ as an infinite series. We recall that the Maclaurin series for the sine function tells us that sin π₯ is equal to the sum from π equals zero to β of negative one to the πth power divided by two π plus one factorial multiplied by π₯ to the power two π plus one for all real numbers π₯. Itβs worth nothing at this point, we can use the Maclaurin series of a function π of π₯ to find the power series representation of a different function.

If we change all instances of π₯ in our Maclaurin series for π of π₯ with some function, letβs say π of π₯. Then we get that π composed with π of π₯ is equal to the sum from π equals zero to β of the coefficients of our original Maclaurin series multiplied by π of π₯ raised to the πth power. And this is valid for all values of the absolute value of π of π₯ less than our radius of convergence in our original Maclaurin series. We can use this to find a power series representation of the function sin of π₯ cubed. Weβll do this by setting our function π of π₯ to be equal to π₯ cubed.

Weβll start with our Maclaurin series for the sine function. And we know that this is true for all real numbers π₯, which is the same as saying that our radius of convergence is β. Now, we change all instances of π₯ in Maclaurin series with π₯ cubed. This gives us the sin of π₯ cubed is equal to the sum from π equals zero to β of negative one to the πth power divided by two π plus one factorial multiplied by π₯ cubed raised to the power of two π plus one. And this is true when the absolute value of π₯ cubed is less than the radius of convergence in our original Maclaurin series which we showed was β. Therefore, our series is true for all real numbers π₯.

We can simplify our summand by noticing that π₯ cubed raised to the power of two π plus one is equal to π₯ to the power of three multiplied by two π plus one, which is just π₯ to the power of six π plus three. So we now have a power series representation of the sin of π₯ cubed. Since the question wants us to find an infinite series representing the integral of the sin of π₯ cubed, weβre going to integrate both sides of our expression with respect to π₯.

Since weβre integrating a power series, we can bring the integral inside our sum. This gives us the sum from π equals zero to β of the integral of negative one to the πth power divided by two π plus one factorial multiplied by π₯ to the power of six π plus three with respect to π₯. Since the value of π does not vary with the value of π₯, our coefficients of π₯ are going to be constants with respect to π₯. And our exponents of π₯ are going to be constants with respect to π₯.

We recall for constants π and π, where π is not equal to negative one, then the integral of π multiplied by π₯ to the power of π with respect to π₯ is equal to π multiplied by π₯ to the power of π plus one divided by π plus one plus a constant of integration π. Since our sum starts from π equals zero, six π plus three will never equal negative one. So our exponents of π₯ will never equal negative one. Therefore, we can use our integral rule to integrate all the terms in our series.

So we add one to our exponent to give us six π plus four and then divide by six π plus four. And instead of adding a constant of integration for each of these terms, weβll combine all of these constants of integration into one constant outside of our sum, which weβll call π.

Therefore, we have shown the integral of the sin of π₯ cubed with respect to π₯ is equal to the sum from π equals zero to β of negative one to the πth power multiplied by π₯ to the power of six π plus four divided by two π plus one factorial multiplied by six π plus four plus a constant of integration π.