### Video Transcript

Use integration by parts to
evaluate the indefinite integral of π₯ times the sin of π₯ with respect to
π₯.

In this question, we are asked
to evaluate the indefinite integral of the product of two functions: an
algebraic function π₯ and a trigonometric function sin of π₯. And this should hint to us that
we could try using integration by parts to evaluate this integral, even if we
were not told to use this method in the question. So, letβs start by recalling
the formula for integration by parts.

It tells us the indefinite
integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus
the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. To apply this to evaluate our
integral, weβre going to need to choose which of the factors in our integrand
needs to be π’ and which one should be dπ£ by dπ₯. So, we need to determine how
weβre going to choose which factor will be π’ and which will be dπ£ by dπ₯.

To do this, we need to know
when weβre using this formula, the most difficult part of this expression to
evaluate is the integral. So, we want to choose our
functions to make this integral as easy to evaluate as possible. And usually, we want to do this
by choosing π’ which makes dπ’ by dπ₯ simpler. However, sometimes, we need to
choose our function π’ such that it cancels when multiplied by part of π£. In either case, letβs look at
our functions π₯ and sin of π₯.

We know when we differentiate
π₯, its degree will decrease. However, if we were to
differentiate the sin of π₯, we would just get another trigonometric
function. This would not be any
simpler. So, weβll set our function π’
to be equal to π₯ and dπ£ by dπ₯ to be equal to the sin of π₯. Now, to apply integration by
parts, weβre going to need to find an expression for dπ’ by dπ₯ and π£.

So, letβs start by finding dπ’
by dπ₯. Thatβs the derivative of π₯
with respect to π₯, which is just equal to one. We also want to find an
expression for π£ from dπ£ by dπ₯. And we can recall π£ will be an
antiderivative of this function. So, we know that π£ is going to
be equal to the indefinite integral of the sin of π₯ with respect to π₯, which
we can recall is negative the cos of π₯ plus the constant of integration. However, we donβt need to add a
constant of integration in this case.

To see this, consider what
would happen if we replace π£ by π£ plus πΆ in our integration by parts
formula. In the first term, we would get
π’ multiplied by the constant of integration πΆ. However, in our second term, we
get πΆ times dπ’ by dπ₯. Negative the integral of this
is negative π’ time πΆ, so these two terms cancel. So, itβs another case of
constants of integration canceling, so we donβt need to include this
constant.

Letβs now substitute our
expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯ into our integration by parts
formula. We get the indefinite integral
of π₯ times the sin of π₯ with respect to π₯ is equal to π₯ times negative the
cos of π₯ minus the indefinite integral of negative cos of π₯ multiplied by one
with respect to π₯. The first term simplifies to
give us negative π₯ times the cos of π₯. And in our second term, we take
out the factor of negative one. This means we add the
indefinite integral of the cos of π₯ with respect to π₯.

Now, all thatβs left to do is
evaluate this indefinite integral. We know that the sin of π₯ is
an antiderivative of cos of π₯. So, the indefinite integral of
cos of π₯ is sin of π₯ plus the constant of integration πΆ. Therefore, we were able to show
the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is negative
π₯ cos of π₯ plus sin of π₯ plus πΆ. And it is worth noting we can
check our answer by differentiating it with respect to π₯ and checking we get π₯
multiplied by the sin of π₯.