Unlike doing the integration by parts problem, I think it’s just a fun problem to see because it’s kind of the example a lot of people use and then sometimes you even have a trick problem that’s giving on the really hard math exam or if you go to calculus competitions like I used to in high school. I was actually not that chaos as a high school student but I have to admit that I was at mathlete anyway this is just a fun integration by parts problem because you actually never have to evaluate the final integral.
Let’s say we want to take the integral and it’s a bit of a classic. You will see this. I won’t be surprise if your math teacher does the same problem for you just to show you integration by parts. Let’s take the integral of e to the x you probably never heard of someone called math from the classic but hopefully, you I will instill in you this love for mathematics and you will also consider this to be a classic problem.
Ex times cosine of x. I think you might already see where I am going with this because these are both functions because ex you can take the derivative, you could take the NI derivative and it still stays to the x cosine of x, you take the derivative. You go to minus sine of x and then you take the derivative again then you get to minus cosine of x and you take the derivative again then you get a plus sine of x right so it’s like a cycle and the same thing happen is when we take the anti-derivative. It’s not as cool as e to the x; it doesn’t stay exactly the same but it cycles every time if you take two anti-derivatives you get back to the negative of itself and if you take the two derivatives, you get back to the negative of itself. So it is also a cool function and I think you can start to see where how integration by parts might be cool here. Whenever I do integration by parts, I always like to assume that this is the g’(x), but e to the x is g’(x) because e to the x literally doesn’t change although we could do this problem the other way, maybe I’ll experiment in doing it the other way. But let’s just assume this is g’(x) and let’s assume that this is f(x). This is g’ because it’s the derivative.
Integration by parts says where we take the original functions g(x) and f(x). So g(x), if this is g’(x) what’s g(x)? What’s the anti-derivative of ex? So this is g(x). I actually took the anti-derivative of it but it is the same exact thing. And then time f(x) and then I want to subtract the integral, the indefinite integral of f’(x) well g(x). This is the same as this which are both the anti-derivative of this although they are all the same. So this is g’(x) and this is g(x) sorry and then I want to take the derivative of f”(x), f’(x) so what’s the derivative of cosine of x? It’s minus sine of x. so sine of x, dx and it’s minus sine of x. so I could put a minus here, I could just say put a minus here and make this minuses cancel out and I'll get a plus here.
The integral of ex cosine of x dx is equal to ex cosine of x plus the integral of ex sine of x dx. I hope I haven’t con fuse you too much. Actually, I try to do some integration by parts problems without ex because it is very hard to keep track of what have done here. Actually, this is the anti-derivative. This is the anti-derivative and this is also the anti-derivative. This is g’(x), this is g(x). So once again, I’m not clear whether we made any progress. You know, we’ve gone from ex cosine of xx to ex sine of x.
Let’s take integration by parts again and see what happens. I must going to write on the right side of the equal sine because this might get a little long. So this equal, some of this is going to write this first part ex, cosine of x plus and now lets’ do integration by parts again. Once again now in this round of integration by parts, this was g(x) but now for this rond I'm going to assume it’s g’(x) which doesn’t really make a difference because when I take the anti derivative to g(x) it stays the same and then I'm assuming that this is f(x).
So integration by parts tells us we take f of x times g of x so I take this function and the anti-derivative of this function. Well, the anti-derivative of this function is once again just ex and then f times that function unchanged times sine of x. And from that, I subtract the integral of the anti-derivative of this or I take the g(x) which is ex and then the derivative of f(x), f’(x) so what’s the derivative of sine of x? Well, it’s cosine of x. Let’s see if you’re forgetting anywhere. It seems like I am just keep budding terms and making it more and more complicated. In order to see if we’re getting anywhere let me just rewrite the whole thing and maybe get rid of this frenzy because it’s just a plus so we can actually get rid of the parenthesis.
So this is the original problem, ex cosine of x dx equals, and now, let me switch back to this color and it equals ex cosine of x and then I can just, this parenthesis doesn’t matter because I am just adding everything in the parenthesis. Ex cosine of x plus ex sine of x minus ex cosine of x dx, now you might think that I arbitrarily switch colors here but I'm right when I rewrote this. But if you look, we might see why I actually did switch colors here. See anything interesting? Exactly. This is the same thing as this just a minus right?
So we are going to do something what I think to be fairly cool. Let’s add this term to both sides of the equation. Let’s take this and let’s put it on to the side of the equation. If I take this and put it on this side of the equation, what happens? I didn’t have two of this on the left side of the equation so that becomes and then I could write it out that’s e to the x cosine of x dx plus because I'm taking this up and I'm putting on that side of the equation. E to the x, cosine of x dx and that’s just the same thing as two times the integral of e to the x cosine of x dx and that equals the this term. I am running out of space, ex cosine of x plus ex sine of x. I know it is really messy but all I have to do now to solve this integral is divide both sides by two and then I'm done. So let me write it out because that’s just very exciting where the home stretch is.
So if I just divide both sides by two, I get and I'm going to try it right so you can see everything. E to the x cosine of x dx equals in all that side, I have ex cosine of x plus ex sine of x over two. I think that’s very neat. I think its neat how integration by parts kind of allows us to do this. We actually never even have to evaluate this integral. We said ow well this integral is just the original problem again and you could think about why that happened because with trick function cycles so we had to do integration by parts twice to kind of get back to where we were before. Then we used that to kind of just solve it without actually having to evaluate the integral.
What I also think is when you look at this solution. It’s kind of neat. The anti-derivative of ex, I actually should never forget the plus C. That would have given me like minus one point on the exam. But it was kindly cool. The integral of ex cosine of x is this expression that’s ex cosine of x plus e to the x sine of x divided by two is the average. It is the average of e to the x cosine of x and e to the x sine of x. I think that’s a very neat property and you might want to graph them and play with them but it’s kind of neat. Hopefully, I have convinced you that this is a classic of a problem and you also find it neat. I'll see you in the next presentation.
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