Evaluate the integrals in Exercises 1–8 using integration by p...

Question

Evaluate the integrals in Exercises 1–8 using integration by parts.

∫ arccos(x / 2) dx

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the integral, the integral of inverse sign of x divided by 3 dx using integration by parts. OK, so it appears for this particular prompt we're simply asked to evaluate the integral that is provided to us by using integration by parts method. So now that we know that we're ultimately evaluating this integral by using the integration by parts method. Our first step that we need to take, based on our prior knowledge on how to use integration by parts, is we need to identify the values of U and DV. So we are going to use, which here's an acronym to help us remember. L I A T E, which stands for logariththic, inverse trigometric, algebraic, trigometric, and exponential, and using this rule, since this is our acronym to help us remember this rule, so using our rule, we will need to choose the inverse trigometric function for you. So what does that mean? So we need to note that you for this particular prompt will be equal to. Inverse sign of. Drum roll X divided by 3. So once again, you will be equal to our inverse trigometric function, which will be U is equal to inverse sin of x divided by 3, and DV will be equal to DX. Our second step now is we need to calculate our values of duu and DV, and to solve for DU, we will need to recall and use the chain rule of derivatives of, let's make a note of inverse sign of x divided by a. Which will mean that DU will be equal to 1 divided by the square root of. 1 minus x divided by 3 in parentheses to the power of 2. Multiplied by 1/3 dx, which will be equal to. One divided by the square root of. 9 minus x to the power of 2 dx, and this is our answer in its most simplified form, so I skipped a step. So essentially when you're doing any sort of solving for integrals, you need to make sure when you're going step by step that you're simplifying each little answer to its most simplified form. So with that said. We now need to note, we need to solve for our value of V, and in order to solve for V, we need to integrate DV, which will mean that V is simply equal to X, so V is equal to X. So, our third step now is at this point, we can recall and apply the integration by parts formula, which the formula states that the integral of UDV. Is going to be equal to u multiplied by V minus the integral of VDU. So with that said, applying our integration by parts formula, we could write that the integral of inverse sin of x divided by 3 d x will be equal to x multiplied by inverse sin of x divided by 3 minus the integral of x divided by the square root of 9 minus x to the power of 2 d x. Fantastic. Our fourth step now is we need to evaluate the remaining integral. So in order to solve for the integral of x divided by the square root of 9 minus x 2. DX we will need to recall and use the substitution method. So let us state that W will be equal to 9 minus x 2. That will mean then that DW will be equal to -2 XDX, which will mean that XDX will be equal to 1/2 DW. Fantastic. Moving right along here. So. When we substitute our information, or rather our known variables back into our integral, we will be able to write that the integral of x divided by the square root of 9 minus x to the power of 2. The X will be equal to the integral of 1 divided by the square root of w. Multiplied by parentheses of 1/22 DW. Which will be equal to 1/2 multiplied by the integral of w to the power of 1/2. DW Fantastic. Which when we simplify our final answer down to its most simple form, we will find that, or rather we can write one more step before we skip a step here, so that 1/2 multiplied by the integral of w to the power of 1/2 dw will be equal to 1/2 multiplied by parentheses of 2 multiplied by w to the power of 1/2. Which will be equal to the square root of 9 minus x 2. Fantastic. And that is our final answer written in its most simplified form. So that means that for our 5th and final step, so step number 5 is our last step, we need to plug in the result from the 2nd integral back into our integration by parts expression, and when we do that, we will find that the integral of inverse sign of x divided by 3 dx will be equal to. X multiplied by. Inverse sign up. X divided by 3. Minus negative the square root of 9 minus x to the power of 2 in parentheses plus c or c denotes some constant variable which will be all equal to x multiplied by inverse sin of x divided by 3 plus the square root of 9 minus x to the power of 2 plus c. And this is our correct and final answer. Hooray, we did it. So going back to look at the problem itself, we can safely say without a doubt that our evaluated integral's final answer is once again x multiplied by inverse sign of x divided by 3 plus the square root of 9 minus x 2 plus c. And that's it, that is our final answer. Thank you so much for watching, hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 1–8 using integration by parts.∫ arccos(x / 2) dx

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Evaluate the integrals in Exercises 1–8 using integration by parts.
∫ arccos(x / 2) dx