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

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SUMMARY

The integral ∫(sec^2 x tanx) dx from 0 to ∏/3 can be approached using integration by parts, leading to the equation I = B - ∫(sec^2 x tanx) dx. This results in the relationship 2I = B, allowing for the conclusion that I = B/2. A direct u-substitution is also applicable for this integral, highlighting the oscillatory behavior common in trigonometric and exponential integrals. This discussion emphasizes the importance of recognizing these patterns in calculus.

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  • Understanding of integration by parts
  • Familiarity with trigonometric identities, specifically sec^2 x and tan^2 x
  • Knowledge of u-substitution techniques in calculus
  • Basic concepts of definite integrals
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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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