Integrating Using Trigonometric Substitution

Normally I use this blog to post things that I am currently studying.  But recently I have discovered that my blog is receiving a good amount of page views via google.  So I from time to time I will be posting things that I feel as though people may need help with.

One such topic that came to mind is integrating using trigonometric substitution.  Over the years I have seen this topic taught by many professors and I have yet to see a way that I have liked, they all used methods of memorization instead of methods of logic. (I don't like to memorize things! It is much easier to remember the logic than formulas!)

I will present an easy to remember method of how to do trigonometric substitution.  So, lets start out with an example..

The trick to these integrals is to relate it to a triangle.  If we look at the Pythagorean theorem, we can put it in the form identical to the form in the denominator.

Thus, we know that the hypotenuse to our triangle (the variable c) is equal to x, and a=1.  So let us construct the triangle from this!

NOTE: The choice of which leg to put "1" is completely up to you. Since both "a" and "b" in the Pythagorean theorem can both be legs, it does not matter where you put it.  It will only change your substitution, not your answer.

So from this we see that..

Now that we have these relationships, lets plug it into our integral.

And that's all there is to it!  It all boils down to simply relating the radical term to a triangle using the Pythagorean Theorem.  You MUST remember to substitute for dx! A common beginner mistake is to forget about dx! You need to find dx by first establishing x, then differentiating it!  Let's do a couple more examples.

So a "tip off" that we need to use trig-sub is seeing the radical term.  We once again relate this to a triangle.  The radical is of the same form as the previous example so we know the hypotenuse must be 3 and one of the legs must be sqrt(5)*x

We can now draw some relationships from this and substitute into the integral.

Let's do one more "typical" example then we will use this for something more exotic.  For this one I will do much less explaining. 

We now relate this to the Pythagorean theorem.

We see from the Pythagorean theorem that one leg of the triangle is equal to 3 and the other leg is equal to x. (The choice of which leg goes where is up to you!)

For this last example, I will show you that you can use this technique in more places than just square roots.  This is something that I actually did at one point.  I couldn't remember the integral of 2^x and I didn't have anywhere to look it up. (Before I had a smart phone!)  So I did this to figure it out.

We can construct a triangle from this.  It need not be in the Pythagorean form.  You can do it in the following way.

From this we can draw trigonometric relations..

Pretty awesome!!

Conclusion:

These examples were some basic trig-sub examples.  They can get more complicated, as in having "x's" in the numerator as well.  But just approach them in the exact same manner and they should turn out okay!

Multipole Expansions

The Dipole:

Let's find the potential at point P due to these two charges.

The Quadrupole:

We now have the following setup..

We will once again assume P is very far away from the charges. It isn't shown, but the bottom line from +Q is r_a, the middle line is r, and the top line is r_b.

The General Case:

We now approach the situation in which we have N number of charges in three dimensional space.  We will use the law of cosines expression just as we did before and the binomial expansion.

Multipole Expansion in Cartesian Coordinates:

Recall that the direction cosines are..

We begin in the exact same manner as before except instead of using a binomial expansion we use a Taylor expansion. (Which is really the same thing!)

The monopole is a scalar, or a Tensor of rank 0. The dipole is a vector or a Tensor of rank 1. And the Quadrupole is a Tensor of Rank 2.