Thank you a lot! I will try to deal with indefinite integrals. Justifications for the definite integral are more complicated. Here is the standard argument, somewhat extended. We can check this by differentiating. One might, as you do, legitimately wonder about this. One point to be made in favour of it is that the above argument proves that the procedure will always give the right answer.
But it all amounts to the same thing. It gets more complicated. But we can check that the symbolic manipulations that we do are still the Chain Rule in disguise, and we can certainly always check whether our symbolic manipulations give the right answer.
Comment: The following general idea is useful. Differentiation is ordinarily easy, integration not so much. If after some work one has calculated an indefinite integral, one can quickly check whether the answer is right, by differentiating. This can save us from errors both major and minor.
Let's check whether this is right. Differentiate, using the Chain Rule. Oops, wrong answer! What you are looking for is an intuitive interpretation of the formula for integration by substitution. I will try to provide it in rather simple words. So substituting alone is not enough. It can be seen from the plot that the areas under the curves differ a lot and, hence, the antiderivatives will differ as well. In the plot, the density reflects this fact: For example, some intervals of the function on the top map to smaller intervals on the substituted function indicating "a higher density".
A high density here means also that some area below the graph gets lost, because the graph becomes "squeezed". Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well. Although we will not formally prove this theorem, we justify it with some calculations here.
So our substitution gives. Use the process from Example to solve the problem. Substitution may be only one of the techniques needed to evaluate a definite integral.
All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in Example. Let us first use a trigonometric identity to rewrite the integral.
We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have. As mentioned at the beginning of this section, exponential functions are used in many real-life applications.
The number e is often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. When the inside function is linear, the resulting integration is very predictable, outlined here. Our next example can use Key Idea 10, but we will only employ it after going through all of the steps.
We can now evaluate the integral through substitution. One may want to continue writing out all the steps until they are comfortable with this particular shortcut. Not all integrals that benefit from substitution have a clear "inside" function. Several of the following examples will demonstrate ways in which this occurs. There is not a composition of function here to exploit; rather, just a product of functions.
The substitution becomes very straightforward:. Our examples so far have required "basic substitution. But at this stage, we have:. Checking your work is always a good idea. In this particular case, some algebra will be needed to make one's answer match the integrand in the original problem. This is another example where there does not seem to be an obvious composition of functions.
The final answer is interesting; the natural log of the natural log. Take the derivative to confirm this answer is indeed correct. Section 6. The next three examples will help fill in some missing pieces of our antiderivative knowledge.
We know the antiderivatives of the sine and cosine functions; what about the other standard functions tangent, cotangent, secant and cosecant? We discover these next. The previous paragraph established that we did not know the antiderivatives of tangent, hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral. We can integrate:. These two answers are equivalent.
This example employs a wonderful trick: multiply the integrand by "1" so that we see how to integrate more clearly. In this case, we write "1" as. This may seem like it came out of left field, but it works beautifully. We summarize our results here. Another technique is needed. The right hand side of this equation is not difficult to integrate. We have:. We'll make significant use of this power--reducing technique in future sections.
It is common to be reluctant to manipulate the integrand of an integral; at first, our grasp of integration is tenuous and one may think that working with the integrand will improperly change the results. Integration by substitution works using a different logic: as long as equality is maintained, the integrand can be manipulated so that its form is easier to deal with.
The next two examples demonstrate common ways in which using algebra first makes the integration easier to perform. When dealing with rational functions i. Hence we use polynomial division. This is very similar to the numerator.
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