Improper Integral of theta/cos^2 theta

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  • #1
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Homework Statement



Improper Integral of theta/cos^2 theta

Homework Equations




The Attempt at a Solution



Hi all, this was one of the few questions on my final today that I just didn't know how to do. I know how to do trig sub, know all my trig identities and know improper integration, but was a bit at a loss for this one.

I could use a half angle for the denominator --> theta / 1/2 [1 + cos(2 theta)] -->

Maybe integrate, and get theta^2 / 1/2 theta + 1/2(sin 2 theta).

I'm sure what I tried was very wrong, but I wanted to make some kind of attempt.

Edit: nevermind, you can't integrate like that.
 
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Answers and Replies

  • #2
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Homework Statement



Indefinite Integral of theta/cos^2 theta

Homework Equations




The Attempt at a Solution



Hi all, this was one of the few questions on my final today that I just didn't know how to do. I know how to do trig sub, know all my trig identities and know improper integration, but was a bit at a loss for this one.

I could use a half angle for the denominator --> theta / 1/2 [1 + cos(2 theta)] -->

Maybe integrate, and get theta^2 / 1/2 theta + 1/2(sin 2 theta).

I'm sure what I tried was very wrong, but I wanted to make some kind of attempt.

Edit: nevermind, you can't integrate like that.
$$\int \frac{\theta d\theta}{cos^2(\theta)} = \int \theta sec^2(\theta) d\theta$$

Use integration by parts with a judicious choice for u and dv.
 
  • #3
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Damn, that's a pretty easy integration by parts question actually. So, if I get an answer of tan(theta) - ln(sec(theta)), where would the improper integration come into play?

Oh wait, natural log functions must be greater than zero. So, it would be something like, the limit, as b approaches 0, from the right, of tan(theta) - ln(sec(theta))?
 
  • #4
34,688
6,394
Damn, that's a pretty easy integration by parts question actually. So, if I get an answer of tan(theta) - ln(sec(theta)), where would the improper integration come into play?
It's not a hard integration by parts, but the answer you show is incorrect. If you differentiate your answer, you don't get ##\theta sec^2(\theta)##.
leo255 said:
Oh wait, natural log functions must be greater than zero. So, it would be something like, the limit, as b approaches 0, from the right, of tan(theta) - ln(sec(theta))?
To be more precise, the argument of a log function must be greater than zero. The output of a log function can be any real number.

The integral you showed was an indefinite integral. An improper integral is a definite integral for which the integrand is undefined at one or more points inside the interval defined by the limits of integration, or at one or both endpoints.
 

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