We learn a new technique, called substitution, to help us solve problems involving integration.
If the functions are differentiable, we can apply the chain rule and obtain If the derivatives are continuous, we can use this equality to evaluate a definite integral. Namely,
On the other hand, it is also true that
Since the right hand sides of these equalities are equal, the left hand sides must be equal, too. So, it follows that
This simple observation leads to the following theorem.
We will apply the Integral Substitution Formula (ISF) to the following example.
Does this integral have the structure so that we can apply the ISF?
We set , so , and note that
So we wrote the integral in a form suitable for application of the ISF. Therefore, we can substitute for and for and write
Now, compute the endpoints:
We can easily evaluate the last integral In summary,
The figure below illustrates both integrals.
Therefore, whenever we are faced with a problem of evaluating a difficult integral, we try to replace it with a simpler one, as long as both integrals represent the same net area.
We can solve a problem like this in a slightly different way. Let’s do the same example again, this time we will think in terms of differentials.
At this point, we can continue as we did before and write
Finally, sometimes we simply want to deal with the antiderivative on its own. We will repeat the example one more time in oder to demonstrate this approach.
As before, we set , and compute , thinking in terms of differentials. Now we see that Hence This does not look right, since we need to find an antiderivative of the function and our result is a function of . We can easily fix this, by simply substituting with . Therefore, Finally, we can apply the SFTOC
With some experience, it is (usually) not too hard to see what to substitute as . We will work through the following examples in the same way that we did for Example ??.
Notice that this example is an indefinite integral and not a definite integral, meaning that there are no limits of integration. So we do not need to worry about changing the endpoints of the integral. However, we do need to back-substitute into our answer, so that our final answer is a function of . Recalling that , we have our final answer Reminder: you can always verify your result by differentiating.
If substitution works to solve an integral (and that is not always the case!), a common trick to find what to substitute for is to locate the “ugly” part of the function being integrated. We then substitute for the “inside” of this ugly part. While this technique is certainly not rigorous, it can prove to be very helpful. This is especially true for students new to the technique of substitution. The next two problems are really good examples of this philosophy.
Then we substitute back into the original integral and solve:
To summarize, if we suspect that a given function is the derivative of another via the chain rule, we introduce a new variable , where is a likely candidate for the inner function. We rewrite the integral entirely in terms of , with no remaining in the expression. If we can integrate this new function of , then the antiderivative of the original function is obtained by replacing by .