Video: Finding the Integral of a Function Using Integration by Parts

Use integration by parts to evaluate ∫ π‘₯ sin π‘₯ dπ‘₯.


Video Transcript

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

The function we’re looking to integrate is the product of two functions. That’s 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 𝑒 times d𝑣 by dπ‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. If we compare this formula to our integrand, we see that we’re going to need to decide which function is 𝑒. And which function is d𝑣 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 𝑒 to be a function that either becomes simpler when differentiated or helps to simplify the integrand when multiplied by the function 𝑣. It should be quite clear that, out of π‘₯ and sin π‘₯, the function that becomes simpler when differentiated is π‘₯. So we let 𝑒 be equal to π‘₯ and d𝑣 by dπ‘₯ be equal to sin π‘₯. The derivative of 𝑒 with respect to π‘₯ is one. And what do we do with d𝑣 by dπ‘₯? Well, we’re going to need to find 𝑣. So we find the antiderivative of sin of π‘₯, which is of course negative cos of π‘₯.

Let’s 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 𝑐. And we see that the solution is negative π‘₯ cos of π‘₯ plus sin of π‘₯ plus 𝑐. Now it’s useful to remember that we can check our answer by differentiating. If we do, we indeed obtain π‘₯ times sin π‘₯ as required.

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