 ###### 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.

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# Tricky Substitutions Involving Radicals - Problem 1

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.

Share

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 try another one of those tricky intervals involving radicals. Here I have the interval of x over a radical x plus 3. So again, the radical function here is a composite function. So my instinct is to substitute for the x plus 3, that’s what I’m going to do.

So w equals x plus 3 and that means dw equals dx. It’s just the derivative of this written in differential form. So this allows me to substitute for the dx here, that becomes dw and the denominator of this radical becomes root w. But what about the x? We got the x left over here. To substitute for the x, I need to use this very same formula and solve it for x.

W minus 3 equals x. I think the reason people resist this is that, they don’t see how it’s going to help. But you have to remember that sometimes making a more complicated substitution for x can actually make the immigration easier even though the algebra extraction is not that different. The immigration will be easier because I can use formulas.

Anyway let’s see how this plays out. The x becomes (w minus 3) dw over root w. Root w is the same as w to the ½. So I’m going to have w over w to the ½, which is w to the ½. It's w to the 1 over w to the ½ w to the 1/2. Minus 3 times w to the -1/2. So this is a much easier interval than what we started with.

These are just powers of w, very easy to anti differentiate using the power formula for antiderivatives. When I want to antidifferentiate this term, I get w to the 3/2 over 3/2 which is the same as 2/3 w to the 3/2. Minus 3 times the antiderivative for w to the -1/2. Again the exponent goes up by 1, -1/2 plus 1 is positive ½. So it's w to the positive ½ divided by positive ½, which is the same as multiplying by 2. Then I add c.

So I have 2/3w to the 3/2 minus 6w to the ½ plus c. All that's left to do is replace w with x plus 3. 2/3 x plus 3 to the 3/2 minus 6 x plus 3 to the ½ plus c. That’s my final answer.