Tricky Substitutions Involving Radicals - Problem 3 2,611 views
When radicals are involved, substitute the expression inside the radical with w. Then, use the substitution method: find dw by taking the derivative of w with respect to x, substitute the x's in the integrand with the expression for w, and integrate as usual. Replace the w's with the expression for x.
Let’s look at a really tougher example that involves integrating radical functions. This one; the integral of 4x² times the square root of 3 minus 2x. Now again, the advice I’ve been giving you is substitute for the thing inside the radical. So we will do that again in here. Then we’ll get w equals 3 minus 2x. And that means that dw equals -2 dx. And I need the dw I need to replace dx with something so I need my dw term here.
However I don’t have a I have a 4 and not a -2. Let me just solve this for dx by, multiplying both sides by -1/2. That will get rid of this -2. So I get dx is -1/2 dw. I’ll use that in a second. So far I’ve got this guy taking care of and this guy but to this one.
For this for x, I need to solve this equation for x. And the way I would do that is I could add 2x to both sides I get 2x equals 3 minus w, and subtract w from both sides and then divide by 2. So x equals ½ 3 minus w. So let’s do that so taking this back into here, I’m going to have the integral of 4 times x².
When I square this, I’m going to get ¼ times this squared. That’s how I’m going to write it in. So times ¼ times 3 minus w² and then I have the radical w then dx is -1/2 dw. So that’s quite a bit there. The 4 and the ¼ cancel and I can pull this -1/2 out in front. But I still have my dw, my root w and my 3 minus w², let me expand this. That’s going to be 9 minus 6w plus w², and this is just w to the ½.dw. See where everything went? Minus ½ is now here, I’ve expanded this binomial and I converted this into an exponent. Now I need to distribute this over each of these terms. My goal is to get just powers of w, nothing but powers of w.
So minus ½, 9 times w to the ½ minus 6, w times w ½ is w to the 3/2. Plus w² times w to the ½. This is the same as w to the 4/2. When you multiply you add the exponents, this will be w to the 5/2. There we have it. Nothing but powers of w very easy to integrate so let me do that.
I still have the -1/2 out in front, I’m going to have 9 times w to the, remember I add 1 to the exponent. So 9 times w to the 3/2 and I divide by 3/2 which is the same as multiplying by 2/3. You can see it’s really important that you have to put something between the 9 and the 2/3 to make sure it doesn’t look something like a mixed number. 9 times 2/3 not 9 and 2/3. Back here, minus 6 times the anti derivative of w to the 3/2. That’s going to be w to the 5/2 divided by 5/2. So minus 6 times 2/5 w to the 5/2.
And finally, anti-derivative of w to the 5/2, I add 1 I get w to 7/2 divided by 7/2 that’s 2/7. So plus 2/7 w 7/2 plus c.
So all I have to do is multiply all these constants out and then replace w with 3 minus 2x. I’m going to the constants first, because this is pretty complicated. So the 9 and the 2/3, the 9 and the 3 is going to cancel leaving a 3. And that makes 6. 3 times 2, 6 times -1/2 is -3w to the 3/2. Then I have 6 times 2 12 over 5 divided by 2, that’s going to be 6 over 5. The negatives cancel, so positive 6 over 5 w to the 5/2. And then minus ½ times 2/7, the 2’s cancel I get minus 1/7w to the 7/2 and I need a plus c still.
So now it’s time to substitute back in 3 minus 2x for w. So I have minus 3 times 3 minus 2x, to the 3/2 plus 6/5. 3 minus 2x to the 5/2. So 5, minus 1/7, 3 minus 2x to the 7/2 plus c. Really complicated looking answer. But this should be the answer to my original anti-derivative of 4x² times the square root of 3 minus 2x.