Evaluate the integrals in Exercises 31–56. Some integrals do n...

Question

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.

∫₀¹/√2 2x arcsin(x²) 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 evaluated from 0 to 1 of x multiplied by inverse tangent of x to the power of 2 dx. OK, so it appears for this particular prong we're asked to ultimately evaluate this integral that is provided to us. So now that we know that we're ultimately evaluating this integral. We need to note that in order to evaluate the definite integral, the integral of, or rather the integral evaluated from 0 to 1 of x multiplied by inverse tangent of x to the power of 2 d x. So if we're going to evaluate this definite integral, we will need to recall and use the substitution method followed by the integration by parts method. So our first step is we're going to recall and apply the substitution method. So let us state that you will be equal to x 2. That will mean that the differential will be du by dx will be equal to 2x, which will mean that x multiplied by dx will be equal to 1/2 multiplied by du. So now, at this point, we need to change the limits of integration. So we need to note when. So let's note that when X is equal to 0, that will mean that U is equal to 0 to 2, which will be equal to 0. And when X is equal to 1, that will mean that U is equal to 12, which will be equal to 1. Which will mean that the integral will become the integral evaluated from 0 to 1 of x multiplied by inverse tangent of x to the 2. The x will be equal to the integral evaluated from 0 to 1 of inverse tangent of u multiplied by 1/2 duu. Which will be equal to drum roll, 1/2 multiplied by the integral evaluated from 0 to 1 of inverse tangent of UDu. OK. So now, our second step is we need to recall and use the integration by parts formula for definite integrals, which states that the integral evaluated from A to B of W multiplied by DV. Will be equal to square brackets of W multiplied by V evaluated from A to B minus the integral evaluated from A to B of VDW. Fantastic. So, keeping our integration by parts formula in mind, let us state that W will be equal to inverse tangent of U, which will mean that dw by DU will be equal to 1 divided by 1 plus u to the 2. Which will mean that dw will be equal to 1 divided by 1 plus u to the power of 2. Do you? So that will mean that DV will be equal to DU, so that will mean that the integral of DV will be equal to the integral of DU, which will mean, without a doubt, that V will be equal to U. So with that said, applying our formula to the integral, we will be able to write 1/2 multiplied by the integral, evaluate from 0 to 1 of inverse tangent of u d u will be equal to 1/2 multiplied by parentheses of square brackets of. You multiplied by inverse tangent. Of you evaluated from 0 to 1, so we need to note that this will be evaluated from 0 to 1. Minus the integral evaluated from 0 to 1 of you divided by 1 plus u to the power of 2DU. Fantastic. So, moving right along here. We can move to our 3rd step now. we now need to evaluate the components, so we now need to evaluate the boundary term and the remaining integral separately. So focusing on the boundary term, which is square brackets of u multiplied by inverse tangent of you. Evaluated from 0 to 1, which will be equal to parentheses 1 multiplied by inverse tangent of 1 minus 0 multiplied by inverse tangent of 0. And let's make a note that since. Inverse tangent of 1 is equal to pi divided by 4. And the second term is equal to 0 based on our prior knowledge because inverse tangent of 0 multiplied by 0 is obviously going to give us a value of 0. We could safely write that square brackets of u multiplied by inverse tangent of you evaluated from. 0 to 1 will be equal to simply pi divided by 4. Focusing our attention now on the remaining integral, we can note that for the integral 0, or rather the integral evaluated from 0 to 1 of you divided by 1 plusu to the power of 2. Do you, we will reckon, well, we should be able to, looking at this integral, we need to be able to recall and recognize this as logariththic form, meaning based on our prior knowledge, we need to recall the integral of Fu divided by Fu is equal to the natural log of the absolute value of f of u. Plusc where c denotes some constant variable. So with that in mind, we can go ahead and write that the integral evaluated from 0 to 1 of u divided by 1 + u to the power of 2 duu will be equal to 1/2 multiplied by the integral evaluated from 0 to 1 of 2 u divided by 1 + u to the power of 2. D u which will be equal to square brackets of 1/2 multiplied by the natural log of the absolute value of 1 + u to the power of 2. Evaluated from 0 to 1. And we need to note, let's make a note that since u to the 2 is greater than or equal to 0, that will mean that u to the power of 2 + 1 is greater than or equal to 1. So that means there is no need for us to write the absolute value sign, so we can eliminate it, we can get rid of it. So we can go ahead and write square brackets of 1 divided by 2 multiplied by the natural log of the absolute value of 1 + u to the power of 2 evaluated from 0 to 1 will be equal to square brackets of 1/2 multiplied by the natural log of parentheses of 1 + u to the power of 2. Evaluated from 0 to 1. Which will be, when we perform our evaluation, will be equal to 1/2 multiplied by the natural log of 1 + 12 minus 1/2 multiplied by the natural log of 1 + 02. Which will be equal to when we perform this calculation, it will be simply equal to the natural log of 2 divided by 2. Fantastic. So now for our 4th and final step is we need to perform our final calculation. So we need to substitute in our results back into our expression from step 2. So we need to take the integral evaluated from 0 to 1 of x multiplied by inverse tangent of x to the power of 2. The x, which is equal to 1/2 multiplied by parentheses of pi divided by 4 minus the natural log of 2 divided by 2. Which will be simply equal to. High divided by 8 minus the natural log of 2 divided by 4, and this is our final answer. Hooray, we did it. So going back to look at the prom itself, we could safely say that our final answer once again. Or our evaluated integral will be as follows. So once again, our integral evaluated from 0 to 1 of x multiplied by inverse tangent of x to the 2 dx is going to be simply equal to. High divided by 8 minus the natural log of 2 divided by 4. 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 31–56. Some integrals do not require integration by parts.∫₀¹/√2 2x arcsin(x²) dx

Watch next

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀¹/√2 2x arcsin(x²) dx