 ###### 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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# Tricky Substitutions Involving Radicals - Concept

Norm Prokup ###### 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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When an indefinite integral involves a function of x times the radical of another function of x, the substitution method must be used in multiple ways. In these tricky substitutions involving radicals, the dx and the x's in both parts of the problem must be substituted for to in order to solve the problems.

Sometimes integrals that involve substitution are very very tricky. Consider this one. the integral of x times radical x+2 dx. How do we solve it? It turns out that a substitution works but it's a very tricky method. So watch how I do it here.
First of all normally when you see a radical, it's a composite function so your instinct might be to substitute for the x+2. And that's a good instinct. You want to substitute for the x+2. It's not a big substitution. The derivative of that is just the differential is just going to be dx. That much will replace the x+2 with a w, so this will become root w. The dx becomes a dw, but you still have an x here, right? I'll draw that in red just to highlight the fact that I've forgotten this guy. So how do you get rid of this part of the integral?
Well, you can use this substitution to solve for x, right? x is going to be w-2, and you can replace the x with w-2. Now first glance, this doesn't look, this doesn't look that much simpler, but it is. It's going to be easier to intergrate. Take a look at what I have here. I can actually distribute the root w over these 2 terms and get 2 powers of w much easier to intergrate than what I started with. w times root w is, remember that root w is like w to the one half. So this will be w to the 3 halves minus 2 times root w. Again, root w is w to the one half. So this is -2w to the one half. Now because these are powers of w I can use the power rule for anti-derivatives and this becomes w to the, I add 1 to the exponent. So 3 halves plus 1 is 5 halves and I divide by 5 halves which is the same as multiplying by the reciprocal, 2 fifths. Minus. I have a 2 here, the anti-derivative for w to the one half is going to be w to the 3 halves because I add 1 to the exponent, w to the 3 halves and I divide by 3 halves which is the same as multiplying by 2 thirds. 2 times 2 thirds. And so my final answer, I should throw a +c in here. My answer is going to be 2 fifths w to the 5 halves minus 4 thirds w to the 3 halves +c. And I still need to re-substitute for w, w is x+2. So my final answer needs to be in terms of x. 2 fifths x+2 to the 5 halves minus 4 thirds x+2 to the 3 halves +c. That's my final answer.
Now the key to this problem was substitution, right? A very simple substitution. I I substituted w for x+2, this little substitution made all the difference because I was able to distribute this root w over the w+, w-2 and get everything in terms of powers of w.
Substitution is all about taking a complicated integral and writing it in a form that allows you to use your integration formulas. That's all that's all that we did here. We just make it possible to use the power rule for anti-differentiation. So that's it. Try tricks like this, this backward substitution in order to make a difficult integral simpler.