Before getting onto the maths proper today I wanted to discuss why you should care about any of this stuff. Who cares whether you can integrate a given function or not? There are many reasons as to why you might care, but I think it’s nice to take an example from either end of the reasoning spectrum.

The first reason is that within many different sciences, the way we describe the world is by differential equations, which we will come on to in the coming weeks. Differential equations are like algebraic equations (that you’ve been studying for years) but they include derivatives. An example is something like:

 

\frac{df(x)}{dx}+x^2+f(x)=2

 

Here you are not being asked to solve for a variable and work out what constant it equals, you are being asked to solve for a function to see how it varies over x.

Such equations are the way we model the world, in just about every field from physics, to sociology and from actuarial sciences to epidemiology. The way we solve such an equation and thus find out about the way a system should change in time and space, is by techniques of integration – as you will see soon. Knowing how to integrate helps you model and study the world in an infinite number of different ways. It is the most powerful tool that science has developed for understanding the universe!

The second reason for studying integration is because it’s a great way to stretch your brain by tackling abstract problems. These problems are like a gym for your brain, and whether or not you go on to work in the sciences, you will have increased your cognitive capacities in ways that someone who has never studied this stuff will not have done. You will be able to understand abstract concepts more quickly and easily than other people, and so this mental workout will stand you in good stead whatever you want to do in the future, and whether or not you ever need to know how to integrate some complicated function.

Right, onto the mathematics…

So, today we started off by clearing up a question which had been raised by something I said during the last lecture. I mentioned that:

 

\int d(anything)=anything+c

 

We can see this intuitively by noting that we are adding up all the infinitesimal pieces of anything – ie. we split it up into infinitesimals, then integrate them, ie. add them all up. What we get back in the end is that anything, plus a constant because we can think of the indefinite integral as the antiderivative.

 

It might be complicating things by calling the variable anything. Instead, if you call it u it might be easier to see, but really it’s just a name, so:

 

\int d(x^3)=x^3+c

 

and the same is true for any function in there. Note that d (x^3)=3x^2 dx and the integral of this returns x^3+c.

 

OK, so we started looking at some more integration by parts examples, so I will write them out here with minor explanations.

 

Example 1

How about the integral:

 

\int(3t+5)\cos\frac{t}{4} dt

 

Well, the first question to ask is which part is going to be simpler if we differentiate it and which is going to be simpler when we integrate it. Clearly differentiating 3t+5 is going to give us a simpler expression, and neither integrating nor differentiating the \cos function is going to change things very much, so we choose:

 

u=3t+5\,\,\,\,\,\,\, dv=\cos\frac{t}{4}dt

 

thus:

 

du=3dt\,\,\,\,\,\,\, v=4 \sin\frac{t}{4}+const

 

Note that in class I always missed out the +const. We will include it here but show that in fact it vanishes from the final expression. Plugging this into our integration by parts algorithm we get:

 

\int(3t+5)\cos\frac{t}{4} dt=(3t+5)(4 \sin\frac{t}{4}+const)-\int (4 \sin\frac{t}{4}+const) 3 dt

 

which we can evaluate to get:

 

(3t+5)(4 \sin\frac{t}{4}+const)+(16 \cos\frac{t}{4}-t const) 3+c

 

Rearranging this gives:

(3t+5)4 \sin\frac{t}{4}+ 48 \cos\frac{t}{4}+5 const+c

 

But we can absorb the const into c to give:

 

\int(3t+5)\cos\frac{t}{4} dt= (3t+5)4 \sin\frac{t}{4}+ 48 \cos\frac{t}{4}+c

In fact whenever you work out the v given dv you can show that you can always ignore that constant of integration, but don’t forget include the constant of integration for the final integral.

Example 2

OK, how about an example where we have to do integration by parts twice. We saw a similar example before, but now let’s look at:

\int x^2 e^{3x}dx

Again, we see that x^2 is going to be simplified in some sense by taking the derivative and so we choose this to be $u$:

u=x^2\,\,\,\,\,\,\, dv=e^{3x} dx

so:

du=2xdx\,\,\,\,\,\,\, v=\frac{1}{3}e^{3x}

Now ignoring the integration constant because we saw how it can be absorbed into the final constant. Plugging u and v into the algorithm, we get:

\int x^2 e^{3x}dx= x^2\frac{1}{3}e^{3x}-\int \frac{1}{3}e^{3x} 2x dx

We still can’t integrate the second term directly yet, but we can apply integration by parts again to this, now choosing u=x and dv =e^{3x}dx (and taking out the constant factors to the front.

Running through the same steps as normal should give you finally:

\int x^2 e^{3x}dx=e^{3x}\left(\frac{x^2}{3}-\frac{2x}{9}+\frac{2}{27}\right)+c

Example 3

In this example we’re going to have a problem which we can solve either by integrating by parts, or by substitution. In general, the more ways you have to tackle the same problem, the more confident you can be with your answers by comparing them.

Let’s look at an integral which at first looks really complicated:

\int \sin x\ln(\cos x) dx

In this case integrating the \ln part gives something more complicated, so we’re going to choose to differentiate it, and thus it will be our u term:

u=\ln(\cos x)\,\,\,\,\, dv=\sin x dx

and thus

du=-\frac{\sin x}{\cos x}\,\,\,\,\, v=-\cos x

so:

\int \sin x\ln(\cos x) dx=-\cos x\ln(\cos x)-\int\cos x\frac{\sin x}{\cos x} dx=-\cos x\ln(\cos x)-\int\sin x dx

which is just

-\cos x\ln(\cos x)+\cos x+c

Now let’s try and tackle the same question with a substitution. Taking our original integral let’s substitute u=\cos x and thus du=-\sin x dx. This gives:

-\int ln u du=-(u \ln u-u)+c=-\cos x\ln(\cos x)+\cos x+c

Which was the answer we got when we used integration by parts.

The last example I gave you to think about was:

\int (\ln x)^2 dx

Give it a go using the same techniques and see what you can get…

How clear is this post?