Integrating x^5sin(x^3) using Integration by Parts - Step by Step Guide

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Homework Help Overview

The problem involves integrating the function x^5sin(x^3) using integration by parts. Participants are exploring the complexities of the integrand and the appropriate choices for u and dv in the integration process.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants suggest using substitution with u = x^3 to simplify the integral before applying integration by parts. Others discuss the LIATE rule for selecting u and dv, questioning its applicability in this case.

Discussion Status

There is ongoing exploration of different strategies for approaching the integral. Some participants have found success with substitution, while others express confusion about the LIATE rule and its effectiveness. Multiple interpretations of the problem are being examined.

Contextual Notes

Participants note the challenge of integrating sin(x^3) and the implications of choosing u and dv from different categories in the LIATE acronym. There is also mention of the original poster's uncertainty regarding the integration process.

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Homework Statement


Integrate x^5sin(x^3). Using Integration by parts.


The Attempt at a Solution



I don't know which one to take du and u
can anyone please help me out?

Thanks a lot before hand!
 
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Well the expressions are pretty complicated and the exponents aren't exactly small if we only get to reduce them one at a time. To simplify a bit and to get familiar sin(u) term instead of the difficult sin(x^3) term try to do the substitution u= x^3 first. Once you have done this you should get an integral that is considerably easier to find using integration by parts.
 
rock_star said:

Homework Statement


Integrate x^5sin(x^3). Using Integration by parts.


The Attempt at a Solution



I don't know which one to take du and u
can anyone please help me out?

Thanks a lot before hand!

Remember the rule known as "LIATE" (aka: Logarithmic-Inverse Trigonometric-Algebraic-Trigonometric-Exponential) the one which comes first is your u and the one which appears later is your dv.

In your question u=x5 because it is algebraic and dv=sin(x3) since it's trigonometric.
 
roam said:
Remember the rule known as "LIATE" (aka: Logarithmic-Inverse Trigonometric-Algebraic-Trigonometric-Exponential) the one which comes first is your u and the one which appears later is your dv.

In your question u=x5 because it is algebraic and dv=sin(x3) since it's trigonometric.

Yea, but that doesn't help, because how will you integrate dv=sin(x3)? rasmhop's excellent advice is what rock_star should follow.

I've never even heard of the "rule" LIATE, and to me, it is more confusing than just understanding how integration by parts works.
 
I agree with you, I didn't notice that :redface: :redface:. I think the OP wanted to find a general rule for choosing u and dv when the integrand is a product of two functions from different categories in the list "LIATE" - in this case you will often be successful if you take the u to be the function whose category occurs earlier in the list and take dv to the rest of the integrand. The acronym LIATE helps one to remember the order but doesn't work in this case, I noticed that can't integrate sin(x^3).

Anyway using rasmhop's substitution did you get: \frac{1}{3} sin(x^3) - \frac{1}{3} x^3 cos(x^3) or is your answer different?
 
Last edited:
roam said:
I agree with you, I didn't notice that :redface: :redface:. I think the OP wanted to find a general rule for choosing u and dv when the integrand is a product of two functions from different categories in the list "LIATE" - in this case you will often be successful if you take the u to be the function whose category occurs earlier in the list and take dv to the rest of the integrand. The acronym LIATE helps one to remember the order but doesn't work in this case, I noticed that can't integrate sin(x^3).

Anyway using rasmhop's substitution did you get: \frac{1}{3} sin(x^3) - \frac{1}{3} x^3 cos(x^3) or is your answer different?

Yes :smile: I solved the problem..
I did it through U substitution...
I took x^3 as U and then solved the problem. The got the above answer you mentioned :smile:
and ILATE helped me with other problems of integration!
Thank you so much! :D
 
n!kofeyn said:
Yea, but that doesn't help, because how will you integrate dv=sin(x3)? rasmhop's excellent advice is what rock_star should follow.

I've never even heard of the "rule" LIATE, and to me, it is more confusing than just understanding how integration by parts works.

Thank you for all the help! :D I solved the problem!
Ya.. I took du and u as u mentioned and it really helped! :D
Thank you so much! :smile:
 

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