common integral substitutions

This writeup presumes knowledge of basic (and some not-so-basic) integration techniques, in particular integration of trigonometric functions.

U-Substitution

This is the most beautiful thing to be able to do when one is integrating. It is overall the least amount of work, and the problem is most elegant. Most other integration substitutions and techniques are a mean to attain this elusive goal. The basic premise is that you have both the original function to some power, as well as it's derivative multiplied by it, as the chain rule would dictate. This only works if the difference between the original function and it's derivative under the integral differ by a constant.

Example:

∫(x²+2)² * 2x dx

First, let us assign u to x²+2. This is because 2x is the exact derivitive of this function. Substituting back in, we now have the following:

∫u²du

u = x² + 2
du = 2x dx

Which, if you have any experience with integration, it is simple to progress to the final solution from here.

Example:

∫(x²+4x+4)^-1 dx

This one is far uglier than the previous, however, it is just as easy with some algebra. First, we need to factor the above polynomial. Thankfully, it is a perfect square trinomial, so we attain

∫(x+2)^-2 dx

As our new integral. Now, we can clearly see a substitution for u being present, and the integral dissolves as follows:

∫u^-2 du

u = x+2
du = dx

Again, a simple integration.

Trigonometric (Trig) Substitution

Closely related to U-Substitution, Trig Substitution is a slightly more complicated technique relying on the properties of right triangles, specifically the pythagorean theorem. Trig Substitution is primarily used when one has one of the three following types of terms in the integral:

sqrt(x²+a²)

sqrt(x²-a²)

sqrt(a²-x²)

In each of these, a represents a constant.

Each of these square roots represent a different trigonometric function which you will set to equal x. These are determined through application of the pythagorean theorem in association with each of the above square roots.

In the case of sqrt(x²+a²), we can see then that, were this the standard pythagorean theorem, the above would equal c :

x²+a² = c²

sqrt(x²+a²) = c

This means that were we to construct a triangle, one leg would be x, one leg would be a, and the hypotenuse would be sqrt(x²+a²). We want to designate the value of x to cleanly equate to a trigonometric function, therefore we will make the leg opposite of θ to be x so that

x = a * tan(θ)

Also, then, we have
dx = a * sec²(θ) dθ, and
sqrt(x²+a²) = a * sec(θ)

The other two roots follow suit, with the hypotenuse being x and a, respectively. We can now use these formulas to solve much more complex integrals.

Example:

∫x²(sqrt(16-x²))^-1 dx

First, we notice that this integral is of the form sqrt(a²-x²), meaning that a will be our hypotenuse. This also means that the following is true (draw the triangle if you wish):

x = 4sin(θ)

dx = 4cos(θ) dθ

sqrt(16-x²) = 4cos(θ)

We then go back to our original problem above and substitute in with the values we now have found, thusly:

∫16sin(θ) dθ

After canceling the 4cos(θ) terms in the numerator and denomerator, of course.

From this step onward, it is a simple matter of using trigonometric identities to produce a solvable integral.

Example:

∫5(sqrt(x²+4x+5))^-1 dx

This one is a bit tricker, as at first glance we don't have a problem of any of the above forms. However, the good friend of completing the square is here to assist us. Using that, we attain the problem:

∫5(sqrt(x²+4x+4-4+5))^-1 dx

Which becomes:

∫5(sqrt((x+2)²+1))^-1 dx

This function now is of the form sqrt(x²+a²), so it is now available for use of trig substitution as follows:

x = tan(θ) - 2

dx = sec²(θ) dθ

sqrt((x+2)²+1) = sec(θ)

Substituting back in, we have the following integral:

∫5cos(θ)sec²(θ) dθ

Which, though not easily solved, is solvable through creative use of trigonometric identites.

With both of these methods, should you have limits of integration, remember to adjust them accordingly for your new variables. For instance, if the upper limit was 3 in our first U-Substitution problem, then we would have to put it into u = x² + 2 to attain our new upper limit (which would be 11, in this case).

Also remember to resubstitute for x in both types in your final answer. This way you can effectively solve a definite integral with these techniques, as well as get the right answer on your Calculus midterm.