What are the limits for these integration problems?

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Homework Help Overview

The discussion revolves around integration problems involving trigonometric and algebraic expressions. The first problem involves integrating a trigonometric identity, while the second problem deals with a rational function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of trigonometric identities to simplify the first integral. There is a discussion about the approach to integrating the second expression, with suggestions to break it into simpler fractions.

Discussion Status

Some participants have provided guidance on how to approach the integration problems, while others express uncertainty about the instructions given by their teacher regarding simplification. The discussion includes attempts to clarify the steps involved in solving the problems.

Contextual Notes

There is mention of specific homework rules that discourage simplification, which may affect how participants approach the problems.

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Homework Statement


A. (1-sin^2x)/(cos^2x) dx
Upper Limit = pi/4
Lower Limit = 0

B. (x-2)/(sqrt x) dx
Upper Limit = 4
Lower Limit = 1



Homework Equations





The Attempt at a Solution


A. I am having trouble with this one.
B. (x - 2)/(x^1/2) = (x^1/2 - 2x^-1/2) =
2/3x^3/2 - 4x^1/2

I plug in 4 and 1 and I get 2/3
 
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For part a, think about the identity [tex]sin^2x + cos^2x = 1[/tex]
 
Oh I KNEW it! That was the only thing I could come up with but I wasn't sure if I could do that.

(1-sin^2x)/(cos^2x) dx

ummm so I get...

(cos^2x)/(cos^2x) dx = 1 dx = x??
 
That's correct, now just plug in the limits. For part b think about breaking up the one fraction into two fractions, each with the same denominator, then you can use the power rule for integrating.
 
A. = pi/4
B. Do you mean like simplify it? If so then our teacher doesn't want us to simplify at all. If not I don't understand what you mean.
 
Ok nevermind, I didn't see that you had already solved it in your first post. It's correct.
 
hotcommodity said:
Ok nevermind, I didn't see that you had already solved it in your first post. It's correct.
Oh okay...well thank you so much for your help! I have a big test on Monday so I am sure I will be on later but I am too tired right now hehe. Thanks again!
 
You're quite welcome, good luck on your test :)
 

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