Integrating by parts(To find integral of product of two functions)

Integrating a product: integration by parts.

Suppose we want to integrate Xsin(X). We can integrate each of these functions separately, but how do we integrate the product of them?
The answer is we use a technique called "integration by parts". First I'll tell you the formula, then show you how it works with that particular example, and then explain where on earth the formula came from!
Here's the method. We have a product of two functions and we want to integrate this product. The first step is, we have to choose one of them to be integrated and one to be differentiated. (The result will be that the whole product gets integrated, but to get there we actually have to differentiate one of them. It seems strange but you'll see why later.)
So we choose one to integrate and the other to differentiate. To remind us which one we're going to integrate we'll write it as the derivative of something, let's call it dv/dX, so when we integrate it we get v(X). (Of course when choosing which function to integrate we have to make sure we can integrate it!)
We'll write the other of the two functions, the one that we're going to differentiate, as u(X).
Before this gets any more confusing, let's see how it works so far on our particular example of Xsin(X).
Here the two functions are X and sin(X). We'll choose the sin(X) to be the one we'll integrate (more later on how to make this choice) and so "X" will be the one we differentiate.
So we write sin(X)=dv/dX and work out the function v(X) that is the integral of this.
Well, the integral of sin(X) is -cos(X), so in this example v(X)=-cos(X).
Now we go on with the method. Having integrated one of our functions, namely dv/dX, to get v(X), we'll now differentiate the other function, which we called u(X), so that'll give us du/dX.
Now we have ended up with four functions altogether: dv/dX, v, u, du/dX. Just plug them into the following formula!
udv/dX dX = uv-vdu/dX dX.
On the left-hand side is the integral we want to do, the product of the two original functions that we called u(X) and dv/dX.
On the right-hand side is the product uv, which we can write down since we worked out v and already know u.
Also on the right-hand side is another integral, this time it's the product of v and du/dX that we're integrating. It doesn't seem like the method has helped us much, since we just end up with another integral, but let's see how it works in our particular example of Xsin(X).
We already worked out that v(X) = -cos(X). Remember that u(X) is the other function we started with, so u(X)=X.
So the derivative du/dX is the derivative of X, which is simply 1.
Now we know all the four functions, so what does the formula give?

Xsin(X) dX = X(-cos(X)) - -cos(X) dX.
On the left-hand side we have our integral of a product that we want to find and on the right-hand side we have the product uv and the integral of vdu/dX.
But since du/dX is just 1, this new integral is much easier to do than the original integral on the left-hand side. In fact we can do it straightaway, we know the integral of -cos(X) is -sin(X).
So for this particular example the formula tells us that the integral of Xsin(X) is -Xcos(X)+sin(X)+c.
Differentiate this to make sure you do end up with Xsin(X). This is a very good habit to get into.
Let's have a quick look back over what exactly happened there. We started with the integral of a product of two functions and this mysterious formula told us we could write our integral as the sum of two terms, one of which was also an integral.
Luckily it turned out to be an easier integral and we found our result. This is the crunch with integration by parts: the new integral must be simpler than the one you started with.
Here's another example, to get you used to the idea. What's the integral of Xe^X?
First step: choose one function to integrate and one to differentiate.
In making this choice, remember that the integral you'll be faced with as a result will be the product of this derivative and integral.
The e^X will just give us back another e^X, whether we differentiate it or integrate it, so it has no preference!
If we choose to integrate the "X", on the other hand,then we'll get something like X², which is more complicated. If we differentiate the "X", we'll get 1, which is nice and simple. So let's do that. We'll integrate the e^x and differentiate the x.
Second step: put it all in the formula.
Remember the formula:
udv/dX dX = uv - vdu/dX dX.
In this case this gives us
Xe^X dX = Xe^X - e^XdX,
so
Xe^XdX = Xe^X -e^X +c.
Finally, where does this formula come from? Very straightforward, it just comes straight from the product rule for differentiating products.
Remember this was:
d[uv]/dX=vdu/dX+udv/dX .
Rearranging gives:
udv/dX=d[uv]/dX-vdu/dX.
Now we integrate all terms in this equation:
udv/dX dX = uv - vdu/dX dx.
And that's our mysterious formula!


Courtesy:To London global university for the best ever explanation I have ever seen.