Integration by parts

In this video, we are going to look at integration by parts. So the definition for integration by parts is if we have 2 functions u and v, they are functions of x, then the integral of u multiplied by dv over dx with respect to x is u multiplied by v subtract the integral of v multiplied by du over dx integrated with respect to x. So we’re going to look at an example now, which is a definite integration. So we want to find the integral between pi and 3 pi of x sine of x with respect to x. So let’s call u, x. And let’s call dv over dx, sine x.

Then we know that du over dx if u is x du over dx will be 1. And if dv over dx is sine x, then v will be negative cosine of x. Because if we integrate sine x it gets us negative cosine of x. So using integration by parts, this is the integral that we want to find. And we can see using the integration by parts formula that we can write this as u multiplied by dv over d x. And then using integration by parts this would be uv, the definite integral between pi and 3 pi, so we have to find it between pi and 3 pi.

Subtract the integral of v multiplied by du over dx between pi and 3 pi. And we have u dv over dx du over dx and v above, so we can just substitute them in, such that uv will be negative x cosine x and v du dx will just be negative cosine of x, because du over d x is 1. And then if we integrate negative cosine of x this will be negative sine of x and we’ve got to subtract negative sine x will be plus sine x. And now we just have to substitute the values in for the definite integral.

One hint is when you’re substituting the values in, of course, make sure your calculator is in radians if you’re using radians in the question. And so when we integrate 3 pi into here cosine of 3 pi will be negative 1. So we would have negative 3 pi multiplied by negative 1 is 3 pi subtract cosine of pi would be negative 1. So negative pi multiplied by negative 1 is negative and would be positive pi, but then we subtract in the lower value. So we get negative pi and when we subtract 3 pi and pi into sine of x this gets us zero. So our final answer for question 1 is 2 pi.

So we’re now going to look at the second example using integration by parts. So it’s again a definite integral and its x divided by the square root of 2x subtract 1. So we’re going to integrate this between 1 and 2. So this time, let’s put that u is x, that would mean that dv over dx would be 1 over the square root of 2x subtract 1, which we can rewrite as 2 x subtract 1 to the power of negative 1/2. And du over dx if u was x du over dx will be 1, and v would be 2x subtract 1 to the power of 1/2. And we would get v by integrating dv over dx.

And we can do this using integration by substitution. So we’re going to use integration by parts again, the definitions there as before. So this is what we’re integrating, and we can rewrite that as integration by parts. So integrating something of the form u which is x multiplied by dv over dx, which is 2x subtract 1 to the power of negative 1/2. And that would be using integration by parts uv determined between 2 and 1 subtract the integral of vdu over dx, integrated between 1 and 2. Now we just need to substitute f for n, so uv would be x multiplied by 2x subtract 1 to the power of 1/2 between 1 and 2.

And v du over dx will just be one lot of 2x subtract 1 to the power of 1/2. And then when we integrate this second integral, using integration by substitution again. Well when we substitute 2 into here we get that that’s 2 square root of 3 and when we substitute 1 in that gets us 1 so it’s 2 square root 3 subtract the 1 with the lower value, so subtract 1, and then we’re going to subtract this integral, which when we use integration by substitution we can find is 2x subtract 1 to the power of 3 over 2, all divided by 3 and we do that between 1 and 2, such that we’ve got our first parts again.

And when we substitute 2 into this one, we get square root of 3. When we substitute one in, we get negative 1/3 so if we subtract negative 1/3 that gets a positive 1/3. And when we type this in a calculator, we would get a square root of 3 subtract 2/3 is the answer to the second example.

So in the comments below one task you could do is create your own question, but this time with an indefinite integral instead of definite, which can be solved by integration by parts. And you may want to answer someone else’s question to practise integration by parts. Thank you for listening.