One Integral - Two Different U-subs

Added: 2026-06-02

[00:00:00]Hi, today we are solving this integral and we're going to be using two different U subs. Let's split it into two. For the first one, I'm going to choose U as 9 - X squared. Then, dU equals -2x dx. My limits of integration becomes instead of a zero, I have 9 - 0, which is 9. Instead of 1, I have 9 - 1, which is 8. Now, I have integral from 9 to 8. The denominator is just square root of U. I see that X dx, it's uh this part. It's almost a dU, so I can say 1 and I can say dU. But, I would love to have -2 because my dU has -2 hiding inside. And over here, I don't have a -2. I have to multiply by -1/2 because -1/2 * -2 x dx is going to give me just x dx. In my second integral, I'm going to choose a different U. I notice this this is similar to an integral of arc sign, which means that U is going to be X over 3. dU now is going to be just 1/3 dx. Instead of 0, I still have 0. 0 / 3.

[00:01:03]Instead of 1, I have 1/3. Factor out the 9. Then, instead of X squared, I have X over 3 squared. That's why I chose U to be equal X over 3. And now, I can write down 9 * 1 - U squared. Notice how square root of 9, that is just 3, which can cancel out with this 3. This 3 comes from 1/3 dx. dU has to be 1/3 dx. I don't have 1/3 dx, so I have to multiply my dU * 3 dx is dU * 3. You can see it from here. This 4 is this 4. It stays.

[00:01:36]Now, I am going to switch the limits of integration so that the smaller number is the lower limit of integration here.

[00:01:42]And that will help me get rid of negative in front of 1/2. My first integral is just a regular power rule.

[00:01:48]And my second integral is going to give me arc sign. All that's left for me to do is plug in numbers and simplify. Arc sign 0 is 0.

[00:01:57]So, this is my final answer.

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