Calculus : Integration

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Trigonometric Integrals

The object of this section is to work out how to do some types of integrals of trig functions, particularly where you have powers of trig functions, but also some others. There are three basic types that we will deal with:

Integrals of the Form .

This situation we have three potential cases:

n is odd.

m is odd.

Both m and n are even.

Note that if both m and n are odd, you can use either of the first two methods.

Integrals of the Form .

There are two cases which are easy to deal with:

n is even

m is odd (and n is at least 1)

m is even

You can use the same sort of trick with trigonometric identities to evaluate integrals of the form [Maple Math] .

Integrals of the Form , or [Maple Math] .

These types of integrals are extremely important, because they often occur when working with Fourier series (an important mathematical tool used widely in Engineering applications).

Fortunately they are not difficult to deal with, simply use the appropriate trig identities:

[Maple Math] .

These are then really easy to integrate.

Trigonometric Substitution

At the moment, we can only do an integral like

if it is a definite integral, and even then only if the geometry is nice that we can recognise it as a part of the area of a circle. An initial attempt to do this would be to suggest a substitution of (or something similar), but you can quickly see that this doesn't work.

Inverse Substitution

It turns out that waht we really need is a sort of reverse substitution rule: rather than writing u as a function of x , we instead try to turn it around, and write x as a function of u . To see how this works, in general, let's say that we want to find:

and we try letting . Then it is not hard to see that , and so we get

When we substitute like this it is called inverse substitution .

Generally speaking, inverse substitution is not a good idea, since it tends to make your integral harder rather than easier. However in some cases we can simplify the integrand after the substitution and get something we can integrate. When we do this we feel very proud for being so sneaky.

The other down-side is that if we manage to find the integral in terms of u , then we have to remember to get back to something in terms of x . To do this we need to find the inverse of the substituted function g(u),

ie. write u as a function of x .

So how does inverse substitution help us? In the example above, if we were to set , (where lies in the interval , so that the inverse is well-defined) then we would find that and

But hey! We can simplify what's under the square root to get

And we learnt how to do this in the last section - use the half-angle formula and get:

= =

Finally, you need to convert back by substituting , getting:

Standard Substitutions

There are three basic tricks like this:

when you have something like , the substitution you try is , and to make the inverse work you need in the interval .

when you have something like , the substitution that you try is , and to make the inverse work you need in the range ( ).

when you have something like , try substituting , and you need to be in the interval [ ) or [ ).

Note that sometimes you are better off substituting = whatever is under the square root sign - these tricks are not always going to work easily, or at all.

Also if instead of a square root, you have a power of or or so on, then you can use the same sort of substitution.

Finding Inverses

Often you will find yourself with final answers which have terms which look like:

or similar (basically you have a trig function applied to some inverse trig function). If you recall the way that inverse trig functions work, you will realise that you can often simplify these terms. Remember that equals the angle of a right-triangle with opposite side x , and hypotenuse a . Pythagoras then tells you that the adjacent side is , and so the cosine of this angle is . So

This sort of analysis will always allow you to simplify such expressions.