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Tricky Substitutions Involving Radicals - Problem 2 2,509 views
When radicals or fractional powers are involved, substitute the expression being raised to the fractional power 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 try a trickier problem. Here I have the integral of x plus 3 over 3 minus x to the 2/3. Now so far, you may have noticed that in the examples I have done, I have substituted for the thing inside the radical each time. Here there is no obvious radical but you have to remember that 3 minus x to the 2/3. That is a radical function.
A fractional power represents a radical. The fraction 1/3 represents the cube root, so this is something squared raised to the cube root of something squared. So I want to substitute for the 3 minus x the way I have been substituting for things inside the radical. So let me try that. W equals 3 minus x. And then I need a dw to replace my dx. So dw is –dx, and that means dx is –dw. I'm going to replace the dx with a -dw. I still have this x plus 3 to deal with. So I would have dealt with this and this but not this. So how do I get x plus 3?
It comes from this formula. I can solve for x I can add x to both sides and subtract w from both side. And I can get x equals 3 minus w. But that’s not x plus 3 x plus 3. X plus 3 I’m basically adding 3 to both sides of this. I would get 6 minus w. So I can replace the x plus 3 with 6 minus w. So let me do that so this x plus 3 becomes 6 minus w.
The dx becomes –dw, and the 3 minus x to the 2/3 becomes w to the 2/3. Just a little trick when you multiply a difference by a -1, you can flip the difference. Because -1 times 6 will be -6. -1 times –w would be +w. So this would be the same as w minus 6, I move the dw over a little bit. So this negative is gone because I just distributed over these two. And this w to the 2/3 I’m going to bring up to the top as w to the -2/3. So just a property of exponents I can bring that up. And then, let me distribute this over these two terms. And after I’ve done that, I’ll have just 2 powers of w that I can integrate really easily.
W times w to the negative 2/3 is w to the 1/3 minus 6 times w to the -2/3. So this is my new integral match easier because it just involves powers of w. First the integral of w to the 1/3. When I integrate this, the exponent goes up by 1 so from 1/3 to 4/3.
W to the 4/3, but then I divide by 4/3 but that’s the same as multiplying by ¾. Then I have a minus 6 and when I integrate w to the -2/3, I raise the exponent by 1. When you add 1 to the -2/3 you get 1./3 . So I get w to the 1/3. And I divide by 1/3. That’s the same as multiplying by 3.
Then I add c. So this is going to become ¾ w to the 4/3 minus 18w to the 1/3 plus c. All that remains to be done is to replace w with 3 minus x. So my final answer is going to be ¾, 3 minus x to the 4/3 minus 18 times 3 minus x to the 1/3 plus c. My final answer.
So remember when you see a radical function inside the integral, or a function with a fractional exponent, replace what’s inside the fraction exponent or the radical with w. That substitution may work for you.