Norm Prokup

**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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The substitution method is useful on some indefinite integrals that are not as simple as they look. These include functions such as e^(5-2x) or the square root of [x-3], in which the variable part of the function is more complex than just an x. These **simple substitutions** require use of the substitution method to solve.

When you're solving indefinite integrals, there are some deceptively simple situations that require the method of substitution. Let's take a look at an example.

First of all, recall the power rule for antiderivativeS. The integral of x to the n is 1 over n plus 1 x to the n plus 1 plus c. Now, when I compare 2 integrals, this really simple integral root x and this slightly more complicated one but still fairly simple root 4x+3. Now when I integrate root x, I can just immediately convert that to a power of x and use the power rule from up here. The question is can I do that for root 4, 4x+3? I will get this answer right? Just two thirds 4x+3 to the 3 halves plus c. Just following the exact same pattern.

Now I want to check this answer by differentiation to see if it's correct. So only take this answer and differentiate it. And I have two thirds 4x+3 to the 3 halves plus c. Okay, the two thirds will come on in front and then fourth, 4x+3 to the 3 halves, the 3 halves will come out and I have 4x+3 to the one half, right? This power minus 1. But I still have to multiply by 4, the derivative of the inside part. Have to use the chain rule. And of course the derivative of the +c is 0. Now when I look at this I get two thirds and 3 halves. This is just 1, then I have a 4 here. 4 times the square root of 4x+3.

Now let's take a look at what we started with. The integral of root 4x+3. We didn't have a 4 in the front. So this method has not worked. In general, it's not going to work unless you have just a single variable raised to a power, right? The fact that I have a 4x+3, this is not the same as x. I have to use the method of substitution in a case like this even though it's a very simple integral.

So integral 2 requires substitution. Let's take a crack at it. So I have the integral of root 4x+3 dx. Remember when I use the method of substitution, I look for a composite function, and this is a composite function. I want to substitute for the inside part. So w=4x+3 and I want to calculate a dw, so the dw is going to be 4 times dx. Now, I have a dx. I don't have a 4 times dx. So what I need to do is multiply both sides by one quarter. So a quarter dw is going to equal dx, and this is what I'm going to replace my dx with. One quarter dw.

So I get the integral of the square root of w times one quarter dw. Now I can switch this to w to the one half and pull the one quarter up in front. So one quarter w to the one half dw. And now I can use the power rule. Here I do have just a simple variable raised to a power. So it's one quarter, I have to raise the exponent by 1. So one half becomes 3 halves and then dividing by 3 halves is the same as multiplying by 2 thirds. This gives me one sixth, right? The twos cancel. I get a one half times one third. one sixth w to the 3 halves plus c.

And the last thing to do is resubstitute for w. I need to put a 4x+3 in there. So this is one sixth sorry. 4x+3, 2 3 halves plus c. That's my final answer. Not exactly the same as the 2 thirds that I thought it was before. This is the correct answer. One sixth 4x+3 to the 3 halves plus c.