Substitution is the most powerful and at the same time perhaps the easiest method of evaluating integrals, which is why it is our method of choice. It has one disadvantage: sometimes it fails. Therefore we not only need to learn the mechanics of substitution itself, but also acquire some feeling for when the substitution works and when it fails. This comes with experience. The general idea of sustitution (not just in integral) is described in this note.

Direct substitution

is based on this mathematical theorem. In the ideal case it can be expressed by this procedure:

That is, we choose a transformation y = g(x) and use it to literally replace all appearances of g(x) with the symbol y. We also need to replace dx. For this we use the equality we deduced in the second row, according to it we replace g'(x)dx with dy.

If this substitution succeeds, we obtain a new integral and with a bit of luck we can solve it. Then one has to also do a so-called back substitution, that is, return back to the original variable, but that is simple, we have a formula for it.

How does this tie in with the formal theorem on substitution? We have a function f(y) = ey and we are interested in in an integral that features f composed with g(x) = sin(x). This more complicated integral is considered on the interval J of all real numbers, since we always want the integral on the largest possible set and there is no reason to restrict ourselves to something smaller. The function g maps this interval onto the interval I = [-1,1]. According to the theorem on direct substitution, it is enough to find an indefinite integral of the function f on the interval I, which is the middle part in the calculation above. Into the formula that we obtained we substitute g and get the answer to our original question, valid on the original interval I.

The beauty of the procedure we use above above is that it allows us to dispose of all this theoretical reasoning, we do not worry about J and I, we simply want to exchange one expression for another. We also need not remember that according to that theorem we also need to have the derivative of g in that complicated integral with composed f. The substituting procedure forces us to replace not dx but g'(x)dx. If we do not find it there (that is, if we did not have that cosine in the integral), then this substitution can't be done. Thus when doing practical calculations people do not worry about that theorem, the procedure we have shown above guarantees that everything is fine.

The main weakness is that we need to have that g'(x) next to dx and that x must appear exclusively in the form of g(x), otherwise the chosen substitution with g would not be possible. However, this is rarely possible. Fortunately, one can help it, which makes substitution one of the most powerful tools. How does it work?

The basic requirement is that after the substitution, all x must disappear from the integral. The only tool available is the chosen transformation y = g(x) and the related equality y = g'(x)dx, but we are also allowed to change these equalities algebraically in order to obtain formulas that allow us to replace expressions featuring x with expressions featuring y. This gives us tremendous freedom, but one should remember that in such a case we do not use the quoted theorem as stated, so then we are on a shaky ground and the answer may come out wrong. It is therefore strongly recommended to check that our result is correct (by differentiating it) - actually, we should do this with every integral we solve. Actually, there is rarely any trouble unless you get really wild with those formulas. We will now show one example of this more general approach.

Linear substitution.
By this we mean any substitution of the form y = Ax + B, which leads to the equality dy = Adx. Here we do not have to rely on the expression Adx appearing in the integral, since we can always rearrange this equality to dx = (1/A)dy and then replace dx in the integral. This means that we can always do any linear substitution. We will show one fairly important example.

The integral is of course only valid for a > 0, otherwise the generalized exponential does not exist. Many people consider this integral an elementary one and memorize it.

Indirect substitution

is based on the following mathematical theorem. In the ideal case it can be expressed by this procedure:

It looks as if we followed the above direct substitution in the opposite direction and there is something to it. There we tried to simplify the integral by replacing a (complicated) expression with a single letter, here we make the integral more complicated, but the basic mechanism stays the same: We choose a transformation and then use it to change the integral, we also get an equality for changing the defferential int he same way, by taking derivative of g and using it as a coefficient. However, deep inside it is a different procedure, which can be seen for instance from the fact that here requirements on g are more strict (in particular, it must be invertible, unlike the situation in direct substitution).

There is also a difference in difficulty of various stages. The direct substitution tight at the start, especially with replacing the dx which often makes the chosen substitution impossible, but the back substitution is clear. On the other hand, the indirect substitution has no trouble with dx, from x = g(t) we immediately get x = g'(t)dt, but the back substitution can be quite an adventure, especially if we want the answer in a nice form.

Now why would anyone want to make an integral more complicated then it already is? Not surprisingly, the indirect substitution is used quite rarely, but there are specific types of integrals where it does help and experienced integrators know beforehand that certain indirect substitutions will in the end magically simplify (see integrals with roots in Methods Survey - Integration). We will show a very simple example of this kind.

Actually, this is an elementary integral that one should remember.

Note that even this simple example is already good enough to illustrate that the indirect substitution is not as simple as it looks. A careful reader should get mightily suspicious at the first integral in the second line: Shouldn't there be an absolute value in the denominator? That is a very good question and the answer depends heavily on how we actually do this indirect substitution. According to the theorem we are supposed to take a function g = sin(t) that goes from some interval J to our interval (-1,1). This interval J must be chosen in such a way that g is 1-1 there and the image of J under g must be the whole interval (-1,1). The most natural choice is the interval , so the new integral, after the substitution, is considered on this interval. There the cosine is always positive, so indeed, there is no need to use an absolute value in the denominator when cancelling the square root and the square power there. Note, however, that if we chose a different interval for J, then we might be forced to use the absolute value. As we see, this is another difference, in direct substitution we need not worry about intervals used in the procedure.

Mixed substitution

is a combination of direct and indirect substitution. We use it to shorten calculations when we have to use several substitutions in a row. Experienced integrators can save quite a lot of time this way. It starts with a transformation h(y) = g(x) and the second important equality is dediced using the usual idea, we differentiate on both sides and attach differentials: h'(y)dy = g'(x)dx. We present one example here that can be also solved by applying two simpler substitutions one after another (try it, it is a good practice):

Also here one has to be careful on which interval the new variable lives, in this particular example one has to decide between intervals [0,) and (-,0], since y2 is supposed to be 1-1 on the working interval. It will not make any difference at the end, but it may strongly influence calculations that get us there, we will see some of it below.

Substitution in a definite integral

This is basically the same as using substitution in an indefinite integral. We state that the basic rule in substitution is to change everything from the language of x to the language of y, and that include also the limits. How do we transform them? As usual, using our basic transformation y = g(x). Example:

Similarly we change limits in the indirect and mixed substitution, but there it is somewhat more difficult. For instance, we had a substitution y2 = x3 + 8 above. If it were a definite integral and one of its limits was x = 2, then we would substitute it into the transformation equation and then look for corresponding y. In this case there are two possibilities, plus and minus 4, the right one is determined by the choice of the interval J in the substitution.

For practical hints and examples see Substitution in Methods Survey - Methods of integration.

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