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 solve another indefinite integral using the method of substitution. I have the integral of 3x over 1 minus 4x². Now here, my usual advice is to look for a composite function. You want to substitute for the inside part of the composite function, but where is the composite function here?
You can think of 1 over 1 minus 4x² as the composite function. That’s 1 over x with this inside it. So this becomes the inside function. 1 minus 4x² and if we make that substitution, we still have to calculate dw/dx. Dw/dx is -8x and our dw will be -8xdx. So this differential will become our new differential. We have to get rid of the dx as part of our substitution. All the x's have to go and we have to replace them with dw's.
What are we going to do about this 3x? Because instead of having -8x times dx we have a 3x times dx. We have to make a little adjustment. Well one thing we can do is, multiply both sides of this equation by -1/8. I get -1/8dw equals xdx. So I have an xdx, I can just write this as the integral of 3 times xdx. 3 times this; -1/8dw, over 1 minus 4x², and that’s w. I can pull the constant, -3/8 out of the integral. What I’m left with is 1 over w, dw. Now remember how to integrate 1 over w, dw. We have this formula over here.
I developed this before. Remember that because the derivative of the natural log of the absolute of x is 1/x, the integral of 1/x dx is the natural log of the absolute value of x plus c. I’m going to use this formula right now. Let’s take it back here.
The integral of 1 over w dw becomes natural log of the absolute value of w plus c. I still have my -3/8 out front. All that’s left to do now is to convert back to x's. So -3/8 (natural log the absolute value of 1 minus 4x² plus c). So this is my antiderivative, my set of antiderivatives for this function.
Let’s check this by differentiating it. Let me take this over here. The derivative with respect to x of -3/8 (natural log, the absolute value of 1 minus 4x² plus c) equals. This -3/8 can come out in front. And we have the derivative of natural log of the absolute value of something. From up here the derivative of natural log of the absolute value of something is just one over that something. So times 1 over 1 minus 4x². But, because this isn’t just x, I have to use the chain rule and multiply by the derivative of this thing. And that’s -8x. The derivative of 1 minus 4x² is -8x.
We have a little cancellation here. The -3/8 and the -8 gives me 3 and that’s the same as 3x over 1 minus 4x². That’s exactly what we started with. So this checks. So going back to our answer, this is the correct answer; -3/8 (natural log, the absolute value of 1 minus 4x² plus c).
Unit
Antiderivatives and Differential Equations