How to evaluate -arctan(cosx) from ∏/2 to ∏

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In summary, to evaluate the given integral, we use the formula (1/a)arctan(u/a)+c and plug in the values from ∏/2 to ∏. This means substituting cos ∏/2 for u and cos ∏ for a. Then, we use the unit circle to find the values for tangent that correspond to these cosine values. We get -1 for cos ∏ and 0 for cos ∏/2. Substituting these values into the formula, we get -arctan(-1) - (-arctan(0)), which simplifies to -(-\pi/4) - (0) = \pi/4. This is the final value of the integral.
  • #1
mathnoobie
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Homework Statement


The direction is the evaluate the integral, this isn't really a calculus issue, it's more of a trig issue.
∫sinx(dx)/(1+cos^2x) from ∏/2 to ∏

Homework Equations


(1/a)arctan(u/a)+c

The Attempt at a Solution


I did all the integrating and ended up at
-arctan(cosx) from ∏/2 to ∏
this is where I'm stuck, I know I'm suppose to use the fundamental theorum of Calculus but I don't know what to do once I plug in ∏/2 and ∏. How do I generate values out of this? Do I draw a triangle? Do I use the unit circle? If so, how would I use it
 
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  • #2
mathnoobie said:

Homework Statement


The direction is the evaluate the integral, this isn't really a calculus issue, it's more of a trig issue.
∫sinx(dx)/(1+cos^2x) from ∏/2 to ∏

Homework Equations


(1/a)arctan(u/a)+c



The Attempt at a Solution


I did all the integrating and ended up at
-arctan(cosx) from ∏/2 to ∏
this is where I'm stuck, I know I'm suppose to use the fundamental theorum of Calculus but I don't know what to do once I plug in ∏/2 and ∏. How do I generate values out of this? Do I draw a triangle? Do I use the unit circle? If so, how would I use it

This seems pretty straightforward.
-arctan(cos([itex]\pi[/itex])) - (-arctan(cos([itex]\pi[/itex]/2)))

What is cos([itex]\pi[/itex])? cos([itex]\pi[/itex]/2)?
 
  • #3
Cos ∏ is -1
Cos ∏/2 is 0 I believe
so then I would take the Arctan(-1)-Arctan(0)
So I would just find on the unit circle where tangent equals -1 and 0, then subtract?
 
  • #4
Right.
 

1. What is the formula for evaluating -arctan(cosx) from ∏/2 to ∏?

The formula for evaluating -arctan(cosx) from ∏/2 to ∏ is:

-arctan(cosx) = -arctan(cos∏) - (-arctan(cos∏/2)) = -0 - (-∏/4) = ∏/4

2. Can -arctan(cosx) from ∏/2 to ∏ be simplified further?

Yes, -arctan(cosx) from ∏/2 to ∏ can be simplified to ∏/4.

3. What is the range of -arctan(cosx) from ∏/2 to ∏?

The range of -arctan(cosx) from ∏/2 to ∏ is from ∏/4 to ∏/4.

4. How can I graph -arctan(cosx) from ∏/2 to ∏?

To graph -arctan(cosx) from ∏/2 to ∏, plot points at (∏/2, ∏/4) and (∏, ∏/4) and draw a straight line connecting them.

5. Is there an easier way to evaluate -arctan(cosx) from ∏/2 to ∏?

There are multiple methods for evaluating -arctan(cosx) from ∏/2 to ∏, including using a calculator or trigonometric identities. Depending on the context of the problem, some methods may be easier than others.

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