Integral Question: ∫(sec^2 x tanx) dx from 0 to ∏/3

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Homework Help Overview

The discussion revolves around the integral ∫(sec^2 x tanx) dx, specifically evaluated from 0 to ∏/3. Participants explore methods of integration, particularly focusing on integration by parts and potential substitutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of integration by parts and the resulting oscillatory behavior observed in the integral. There is mention of trigonometric identities that may relate sec^2 x and tan^2 x, as well as a suggestion of a direct substitution method.

Discussion Status

The discussion is ongoing, with participants sharing insights about the integration process and the implications of their approaches. Some guidance has been offered regarding the oscillatory nature of the integral and the importance of recognizing this behavior in similar problems.

Contextual Notes

Participants note the relevance of understanding integration techniques and the potential for encountering similar questions in testing scenarios. There is an emphasis on the importance of recognizing patterns in integration results.

LocalStudent
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Homework Statement



∫(sec^2 x tanx) dx

the integral is from 0 to ∏/3

Homework Equations



I tried using integration by parts


The Attempt at a Solution



When I did integration by parts I got to this:

tan^2 x - ∫(sec^2 x tanx) dx (integral also from 0 to ∏/3)

which take you back to the question I started with.
 
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Hint: Are there trigonometric identities that relate sec^{2}x and tan^{2}x?
 
LocalStudent said:
When I did integration by parts I got to this:

tan^2 x - ∫(sec^2 x tanx) dx (integral also from 0 to ∏/3)

which take you back to the question I started with.
There is a direct u-sub to handle this integral, but the way you have done it does illustrate an important concept.

Suppose, I = ∫Adx and you perform integration by parts and get I = B - ∫Adx. Well, that second integral is, as you noted, what you started with. Thus, I = B - I and so 2I = B and thus, I=B/2.

That type of oscillatory behavior is very important for many trigonometric and exponential integrals.
 
Jorriss said:
Suppose, I = ∫Adx and you perform integration by parts and get I = B - ∫Adx. Well, that second integral is, as you noted, what you started with. Thus, I = B - I and so 2I = B and thus, I=B/2.
integrals.

Thanks, that was really helpful. I hope I remember and notice it in my test.
 
Jorriss said:
There is a direct u-sub to handle this integral, but the way you have done it does illustrate an important concept.

Suppose, I = ∫Adx and you perform integration by parts and get I = B - ∫Adx. Well, that second integral is, as you noted, what you started with. Thus, I = B - I and so 2I = B and thus, I=B/2.

That type of oscillatory behavior is very important for many trigonometric and exponential integrals.


Thanks again! This exact question came up in my math test :)
 

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