### Video Transcript

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

The function weβre looking to
integrate is the product of two functions. Thatβs π₯ and sin π₯. This indicates that we might need
to use integration by parts to evaluate our integral. Remember the formula here says the
integral of π’ times dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’
by dπ₯. If we compare this formula to our
integrand, we see that weβre going to need to decide which function is π’. And which function is dπ£ by
dπ₯. So how do we decide this? Well, our aim is going to be to
ensure that the second integral we get over here is a little simpler. We therefore want π’ to be a
function that either becomes simpler when differentiated or helps to simplify the
integrand when multiplied by the function π£. It should be quite clear that, out
of π₯ and sin π₯, the function that becomes simpler when differentiated is π₯. So we let π’ be equal to π₯ and dπ£
by dπ₯ be equal to sin π₯. The derivative of π’ with respect
to π₯ is one. And what do we do with dπ£ by
dπ₯? Well, weβre going to need to find
π£. So we find the antiderivative of
sin of π₯, which is of course negative cos of π₯.

Letβs substitute what we have into
our formula. We see that our integral is equal
to π₯ times negative cos of π₯ minus the integral of negative cos of π₯ times one
dπ₯. This simplifies to negative π₯ cos
of π₯. And then we take the factor of
negative one outside of our integral. And we see that we add the integral
of cos of π₯ evaluated with respect to π₯. The antiderivative of cos of π₯ is
sin of π₯. And of course, because this is an
indefinite integral, we must add that constant of integration π. And we see that the solution is
negative π₯ cos of π₯ plus sin of π₯ plus π. Now itβs useful to remember that we
can check our answer by differentiating. If we do, we indeed obtain π₯ times
sin π₯ as required.