Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
To solve an integral with the substitution method, look for a composite function. In most cases, the function containing the other function will be what you substitute (with u or w, whichever variable is preferred. Here, we will use w). When using the substitution method, you must turn all x's into w's. So, when you define w, solve for x; then, you will be able to replace x's in any other part of the integrand with an expression containing w's. For example, if w = 2x + 3, then x = (w - 3)/2. In addition, the dx must be converted to dw. To do this, simply take the derivative of w with respect to x. Continuing with the previous example, if w = 2x + 3, dw/dx is 2. So dw = 2dx, and dx = ½dw.
After substituting all of the x's in the integrand with the appropriate expressions with w, then integrate as usual. Remember to replace the w's in the result with the expression of x that was substituted.
Let’s try a harder indefinite integral problem. I want to integrate x² plus 2x over the square root of x cubed plus 3x² plus 1. That’s a pretty complicated looking function but it turns out that the method of substitution will help us out really nicely here. Remember substitution is just a change of variables to make the integral simpler.
Well remember, what I always look for is a composite function. I see one here. And I always try to substitute for the inside part of the composite function. So I’m going to substitute for x cubed plus 3x² plus 1. That’s the inside part. The derivative of that; dw/dx, that’s going to be 3x² plus 6x.
Now, when I look at what I have here, I have an x² plus 2x, it’s very similar. I’m going to have to maybe pull a 3 out of this. And I also need to multiply both sides by dx. So I need dw equals 3 times x² plus 2x times dx. I’ve just pulled a 3 out of each of these two terms, leaving the x² plus 2x and I’ve got my dx for the left side. This is how I’m going to substitute for dw.
Notice, if I divide both sides by 3 I get (x² plus 2x) dx, that’s exactly what I have on the numerator. So my numerator is going to become 1/3dw and my denominator is the square root of w. That is a much simpler integral.
Now I need to express this in terms of a power of w. We pull the 1/3 out. 1 over the square root of w is w to the -1/2. And now, this allows me to use the power rule for antiderivatives. The antiderivative for a power like this is w to the -1/2 plus 1, over -1/2 plus 1. I have the 1/3 in front and the plus c at the end.
Now -1/2 plus 1 is ½, so that’s ½ here and here. Dividing by ½ is the same as multiplying by 2, so I get 2/3w to the ½ plus c. All that’s left to do is to plug in where the w originally was, which is x³ plus 3x² plus1. My answer is 2/3x³ plus 3x² plus 1, and that’s to the ½ plus c. Now let me check this answer by differentiating. Let me take it up here.
The derivative with respect to x, of 2/3x³ plus 3x² plus 1 to the ½ plus c. At first, the constant can be pulled right out. I get 2/3 times, and the derivative of something to the ½, is ½ times that something to the -1/2. The exponent goes down by ½ so I have x³ plus 3x ² plus 1. But I also have to multiply by the derivative of the inside part. That’s the chain rule. So I multiply by the derivative of this which is 3x² plus 6x. And of course, the derivative of the plus c is 0. So this is going to simplify a lot.
The 2's cancel giving me a 1/3 and 1/3 times this, gives me x² plus 2x. So I’ll just write this as x² plus 2x times x³ plus 3x² plus 1. This is exactly the same as what we started with. Just taking you back to where we started. Here is the x² plus 2x. Here is the x³ plus 3x² plus 1 to the -1/2. Our answer does indeed check and it’s 2/3 x³ plus 3x² plus 1 to the ½ power plus c.
Unit
Antiderivatives and Differential Equations