How Do You Evaluate the Integral of arcsec(x) from sqrt(2) to 2?

In summary, there was a conversation about evaluating the integral from sqrt(2) to 2 of 1 over x*sqrt(t^(2) - 1) dx. The person noticed that it was just the arcsec and got arcsec(x) for the answer. They asked how to evaluate this at 2 and sqrt(2), and wondered if there were 2 variables in the integral and if t could be treated as a constant. Further clarification and work were requested. The integral was then rewritten as 1 over x*sqrt(x^(2) - 1) dx, and the person evaluated it to be arcsec(x). However, this didn't make sense with the limits of integration. Suggestions were given to use
  • #1
frasifrasi
276
0
For the integral from sqrt(2) to 2

of
1 over x*sqrt(t^(2) - 1) dx

I noticed that this was just the arcsece, so I got arcsec(x) for the answer, but how would I evaluated this at 2 and sqrt(2)?


What did i do wrong?


Thank you!
 
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  • #2
Why are there 2 variables in your integral? Is t supposed to be there? Can it be treated as a constant for this question?
 
  • #3
more clarity and a little more work would be appreciated
 
  • #4
Ok, the integral is:

1 over x*sqrt(x^(2) - 1) dx


--> which I evaluated to be arcsec (x),but this doesn't make sense with the limits of integration...
 
  • #5
Well if you want to find arcsec([itex]\sqrt{2}[/itex]) you can always work it out like this:

Let [itex]\alpha=sec^{-1}(\sqrt{2})[/itex]

so that [itex]sec\alpha=\sqrt{2}[/itex]
and therefore [itex]cos\alpha=\frac{1}{\sqrt{2}}[/itex] and then you find [itex]\alpha[/itex]OR...somewhere in you attempt you would have used the substitution x=sec[itex]\theta[/itex] so from there you could have gotten [itex]\theta=cos^{-1}(\frac{1}{x})[/itex] and use that instead of arcsec
 
  • #6
Oh my god!
 

FAQ: How Do You Evaluate the Integral of arcsec(x) from sqrt(2) to 2?

How do you solve an urgent trigonometric integral?

To solve an urgent trigonometric integral, you must first identify the function and its boundaries. Then, you can use various trigonometric identities and integration techniques to solve the integral. It is important to be familiar with these methods and have a strong understanding of trigonometric functions in order to solve the integral efficiently.

What is the importance of solving an urgent trigonometric integral?

Solving an urgent trigonometric integral is important because it allows us to find the area under a curve, which has many practical applications in fields such as physics, engineering, and economics. It also helps us understand the behavior of trigonometric functions and their relationships with each other.

Can you provide an example of solving an urgent trigonometric integral?

Sure, here is an example of solving the integral from √2 to 2 of sin(x)dx. First, we can use the identity sin^2(x) + cos^2(x) = 1 to rewrite the integral as ∫(√1-cos^2(x))dx. Then, we can use the substitution u = cos(x) and du = -sin(x)dx to simplify the integral to -∫√1-u^2 du. This can be solved using the formula ∫√1-u^2 du = (sin^-1(u) + u√1-u^2) + C. Plugging back in for u and simplifying, we get -sin^-1(cos(x)) - cos(x)√1-cos^2(x) + C.

Are there any common mistakes when solving an urgent trigonometric integral?

Yes, some common mistakes when solving an urgent trigonometric integral include forgetting to use trigonometric identities, incorrect substitution, and forgetting to add the constant of integration. It is important to double-check your work and make sure you are using the correct methods and identities.

How can I improve my skills in solving urgent trigonometric integrals?

The best way to improve your skills in solving urgent trigonometric integrals is through practice and understanding the underlying concepts. Make sure to review trigonometric identities and integration techniques, and try solving a variety of integrals to get more comfortable with the process. You can also seek guidance from a tutor or teacher if you are struggling with a particular integral or concept.

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