9–40. Integration by parts Evaluate the following integrals us...

Question

9–40. Integration by parts Evaluate the following integrals using integration by parts.

26. ∫ t³ sin(t) dt

Accepted Answer

Hello there. Today we're 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 Y to the power of 3 multiplied by cosine of YDY using integration by parts. Awesome. So it appears for this particular prom we're asked to take our integral that is provided to us by the prom itself and we're asked to evaluate it using integration by parts. So now that we know that we're ultimately trying to evaluate this particular integral using integration by parts, our first step that we need to take in order to solve this problem is we need to recall and use the acronym Lippi. To help us choose the values of U and DV, so this acronym Lipi, which stands for L. I P E T, where L stands for logamithic functions such as the natural log of X, where I denotes inverse trigontric functions such as inverse tangent of X, where P denotes polynomials such as X to the power 2 or Y. E stands for exponential, which is, for example, E to the power of X, and finally T stands for trigontric such as sin of X or cosine of X, etc. So using our acronym Lipi, we will determine when we look at our integral, which is the integral of Y to the power of 3 multiplied by cosine of Y D Y, we will determine that Y to the power of 3 is a polynomial. So we will determine that you, since we're going to choose U to be Y to the power of 3. And we need to also recognize that cosine of Y is a trigonometric function. So we will determine that DV is equal to cosine of Y. DY So that will mean that we can proceed with U equals the power of 3, so we can state that DU is equal to 3 to the power of 2DY, and we can state that DV is equal to cosine of YDY, which we can go ahead and say that V is equal to sin of Y. Fantastic. So if V is equal to sin of Y, for DV is equal to cosine of Y D Y. And we've established our values for you and DV. Our second step now is we need to perform our first integration by parts, and let us recall that the equation for integration by parts, or rather the formula, is the integral of UDV is equal to U multiplied by V minus the integral of VDU. Fantastic. So moving right along here. We now need to apply the integration by parts to our problem, and when we do, we will find that the integral of Y to the power of 3 multiplied by cosine of Y D Y is equal toy to the power of 3 multiplied by sin of Y minus the integral of 3Y to the power of 2. Multiplied by sine of Y D Y, fantastic. So, we now need to compute the integral of 3 to 2 multiplied by sine. Of Y D Y, and in order to do that, we need to factor out the 3. So when we do, we will find that it's equal to 3 multiplied by the interval of Y to the power 2 multiplied by sine of Y. DY Fantastic. So now our 3rd step is we need to perform integration by parts for the 2nd time, except now we are integrating the integral of Y to the power of 2 multiplied by sign of Y. Dy fantastic. So with that in mind, since now we're taking and we're integral the we're integrating the integral of Y, so we're taking the integral of Y squared multiplied by sin of Y D Y, which will mean that we need to state that U is equal to Y2d for this particular case, which will mean that DU. is equal to 2 Y D Y and that DV is equal to sin of Y, which will mean that V is equal to negative cosine of Y. Fantastic. So with that in mind, we can go ahead and write that the integral of Y 2 multiplied by sin of Y D Y is equal to negative Y 2 multiplied by cosine of Y plus the integral of 2 Y multiplied by cosine of Y D Y. Fantastic. So now we need to compute the interval of 2 Y multiplied by cosine of YDY. So, that will mean that we need to, for our 4th step, we need to Perform a third integration by parts, so we need to say that U is equal to Y, which will mean that DU is equal to DY. So DU is equal to DY, and that will mean that DV is equal to cosine of Y, which will mean Once again, that DV is equal to cosine of Y D Y, so that will mean that V is equal to sin of Y. Fantastic. So that will mean drum roll, that the integral of 2 Y multiplied by cosine of Y. and then we're going to take the integral. Once again, the integral of 2 Y multiplied by cosine of Y D Y is equal to 2 multiplied by parentheses Y multiplied by sin of Y minus the integral of sine of Y D Y. Will be equal to 2 multiplied by parenthesis Y multiplied by sign of Y. Plus cosine of Y, fantastic. So now that will mean that our 5th step is we need to put everything together and we need to go back to our second integration. So we need to take the integral of Y to the power 2. Multiplied by sin of Y D Y is equal to negative to the power to cosine of Y plus 2y multiplied by sign of Y plus 2 multiplied by cosine. Of Y. So that will mean that the interval of 3 to the power 2 multiplied by sin of Y. DY So Once again, we're taking the integral of 3 to the power 2 multiplied by sin of Y D Y is equal to 3 multiplied by parentheses negative squared multiplied by cosine of Y plus 2 Y multiplied by sin of Y plus 2 multiplied by cosine of Y. Fantastic. So that will mean. That We can say that this is all equal to -3 2 multiplied by cosine of Y plus 6 Y multiplied by sin of Y plus 6 multiplied by cosine of Y. So now, our 6th and final step is we need to go back to the original integral. So we're gonna be, once again, we're trying to put everything back together. So that will mean that we'll take the integral of Y to the power of 3 multiplied by cosine of Y D Y is equal to the power of 3 multiplied by sin of Y minus -3Y 2 multiplied by cosine of Y plus 6 Y multiplied by sin of y. Plus, 6 multiplied by cosine of Y. So that will mean that this big old long expression will be equal to Y 3 multiplied by sin of Y + 332 multiplied by cosine of Y minus 6 Y multiplied by sin of Y minus 6 multiplied by cosine of Y plus C. And that big old long expression is our final answer. Hooray, we did it. So, going back to the top to look at the problem itself and to summarize what we've learned, that will mean that our final answer once again. Without a doubt, will have to be the following. Which is Y to the power of 3 multiplied by sin of Y plus 3 Y 2 multiplied by cosine of Y minus 6 Y multiplied by sin of Y minus 6 multiplied by cosine of Y plus C. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

9–40. Integration by parts Evaluate the following integrals using integration by parts.26. ∫ t³ sin(t) dt

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9–40. Integration by parts Evaluate the following integrals using integration by parts.
26. ∫ t³ sin(t) dt