What is the antiderivative of (1-cos(t/6))^3 * sin(t/6)?

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Homework Help Overview

The problem involves finding the antiderivative of the expression (1-cos(t/6))^3 * sin(t/6). This falls under the subject area of calculus, specifically integration techniques.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts various methods including integration by parts, trigonometric substitution, and u-substitution but expresses difficulty in finding a solution. Some participants suggest specific substitutions, such as u=(1-cos(t/6)), while others mention reducing the expression to a simpler trigonometric form.

Discussion Status

Participants are exploring different approaches to the problem, with some offering guidance on potential substitutions and methods. There is no explicit consensus on the best approach, but suggestions for simplification and alternative methods are being discussed.

Contextual Notes

The original poster indicates a struggle with the problem despite trying multiple techniques, suggesting a possible lack of clarity on the appropriate method to apply. There may be assumptions about the familiarity with integration techniques that are being questioned.

cemar.
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Hey guys!
So i have to find the antiderivative of
(1-cos(t/6))^3 * sin(t/6)
I have tried a bunch of different ways like by parts, trig substitution, identities, u-substitution then developing it, etc. but i just can't get it!
Can some one just tell me the proper method to use after that I am sure i can figure it out.
thanks!
 
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You must not have tried u substitution very hard. Try u=(1-cos(t/6)).
 
It should be doable. Or so says Wolfram's online integrator. I tried it out and managed to reduce it to a doable trigo form. You only need use integration by parts and then replace all those trig terms which are of power 2 and 3 with power 1 terms by means of some trigo formulae.
 
Defennnder said:
It should be doable. Or so says Wolfram's online integrator. I tried it out and managed to reduce it to a doable trigo form. You only need use integration by parts and then replace all those trig terms which are of power 2 and 3 with power 1 terms by means of some trigo formulae.

I don't see why you are working so hard on it. There's a much simpler way.
 
You're right, I just didn't see it before you posted.
 

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