Evaluating Trigonometric Integral

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Homework Help Overview

The discussion revolves around evaluating a trigonometric integral involving the function sin(y) raised to a power, specifically from 0 to π/2. Participants are exploring integration techniques and the implications of variable substitution on the bounds of the integral.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration process and the use of u-substitution. There are questions about the correct bounds after substitution and the evaluation of the integral. Some participants express uncertainty about their calculations and seek clarification on specific steps.

Discussion Status

The discussion is active, with participants providing feedback on each other's approaches. Some guidance has been offered regarding the adjustment of bounds after substitution, and there is acknowledgment of mistakes made during the evaluation process. Multiple interpretations of the integral's evaluation are being explored.

Contextual Notes

There is mention of a discrepancy between the participant's calculated result and the expected answer from a textbook, indicating potential misunderstandings in the evaluation process. Participants are also addressing the need to clarify the bounds of integration.

jdawg
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Homework Statement


\int\frac{\pi}{2}0 sin7y dy
The bounds are from \frac{\pi}{2} to 0.

Homework Equations





The Attempt at a Solution


I think I did the integration correctly, but I don't really know how to evaluate this.

\intsiny(sin2y)3 dy
\intsiny(1-cos2y)3 dy
u=cosy
du=-siny dy

-\int(1-u2)3 du
-\int(-u6+3u4-3u3+1) du

\frac{u<sup>7</sup>}{7}-\frac{3u<sup>5</sup>}{5}+\frac{3u<sup>4</sup>}{4}-u

\frac{cos<sup>7</sup>y}{7}-\frac{3cos<sup>5</sup>y}{5}+\frac{3cos<sup>4</sup>y}{4}-cosy

I'm a little rusty on my trig! Please help :)
 
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jdawg said:

Homework Statement


\int\frac{\pi}{2}0 sin7y dy
The bounds are from \frac{\pi}{2} to 0.

Homework Equations





The Attempt at a Solution


I think I did the integration correctly, but I don't really know how to evaluate this.

\intsiny(sin2y)3 dy
\intsiny(1-cos2y)3 dy
u=cosy
du=-siny dy

-\int(1-u2)3 du
-\int(-u6+3u4-3u3+1) du

\frac{u<sup>7</sup>}{7}-\frac{3u<sup>5</sup>}{5}+\frac{3u<sup>4</sup>}{4}-u

\frac{cos<sup>7</sup>y}{7}-\frac{3cos<sup>5</sup>y}{5}+\frac{3cos<sup>4</sup>y}{4}-cosy

I'm a little rusty on my trig! Please help :)

No the bounds are not from ##\pi/2## to 0, they are from 0 to ##\pi/2##. Anyway, you have the right idea; you just need to figure out the bounds after the u-substitution; that is, what are ##a## and ##b## in
\int_0^{\pi/2} \sin^7 y \: dy = -\int_a^b (1-u^2)^3 \, du?
 
Oops! That's what I meant. Haha thanks, I guess it would be a lot easier to just change the bounds.
 
Hey! I ran into another problem, I changed the bounds to 1 to 0:

-∫ (-u6+3u4-3u3+1) du

Then I flipped the bounds to 0 to 1 and changed the sign on the integral:

∫(-u6+3u4-3u3+1) du

Then I plugged in my bounds and got 99/140, but the answer in the back of the book is 16/35. I'm having trouble figuring out what I did wrong.
 
jdawg said:
Hey! I ran into another problem, I changed the bounds to 1 to 0:

-∫ (-u6+3u4-3u3+1) du

Then I flipped the bounds to 0 to 1 and changed the sign on the integral:

∫(-u6+3u4-3u3+1) du

Then I plugged in my bounds and got 99/140, but the answer in the back of the book is 16/35. I'm having trouble figuring out what I did wrong.

How do you get ##u^3##? It should be ##u^2## instead.
 
Pranav-Arora said:
How do you get ##u^3##? It should be ##u^2## instead.

Ooops! It was just a silly mistake. Thanks so much!
 

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