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Evaluate the integral or state that it diverges.40 tan 1 + 25y-Select the correct choice below and if necessary; fill in the answer box to complete your choice_40 t...

Question

Evaluate the integral or state that it diverges.40 tan 1 + 25y-Select the correct choice below and if necessary; fill in the answer box to complete your choice_40 tan 1 + 25y2(Type an exact answer in terms of I) The integral diverges.

Evaluate the integral or state that it diverges. 40 tan 1 + 25y- Select the correct choice below and if necessary; fill in the answer box to complete your choice_ 40 tan 1 + 25y2 (Type an exact answer in terms of I) The integral diverges.



Answers

In Problems $31-52,$ evaluate the given improper integral or show that it diverges.
$$ \int_{0}^{\pi / 4} \frac{\sec ^{2} \theta}{\sqrt{\tan \theta}} d \theta $$

We're going to evaluate the improper integral or show that it diverges. So we are integrating from zero to pi over four and we have a denominator of tan x minus one squared. But if the tan x minus one ever equaled zero, that would mean that we were approaching a vertical ascent ope. So where is tan X equal to one? Well, it happens to be at pi over four. So we do have vertical ascent up at pi over four. So instead of integrating from zero to pi over four, we're actually going to integrate from zero to be. We're going to take the limit as B approaches pi over four from the negative side. Okay, so now that we know how to place this as a limit, we need to think about how we're actually going to integrate this. Well remember you can have a you wherever that you needs to be but you can only have a D. U. In the numerator. And seeking squared is the is the derivative of tan X. So to clean our bottom up the most we can we're going to actually make are you equal to Tan X -1? Which will make our D. U. Equal to seek in squared X. Dx. So the whole top will become D. You the bottom becomes U. Squared. So to actually integrate that we want to make it a. U. To the negative too. So now that we have what we're going to integrate in terms of you we can use power rule. So if we go up a power will be going to negative one. And then we have to multiply by the reciprocal which is negative. So as we place this back into our limit we know that we have a negative and we can do that as a one over R. Tan X -1. And that's really just to the first power. So you don't have to write that. Okay, bringing R. B in first and then we're going to subtract bringing in zero. So as we put be in would just replace the X. Would be. Now as we put zero in the tangent of zero is Um actually zero. So 0 -1 would give us a negative one. Okay so now we have to place that pile over four in. For be well tangent of pi over four is one. So we do have one minus one. So we have a divide by zero. So that tells us that we are actually going towards infinity and infinity plus one is still infinity. And so this improper integral diverges.

To evaluate the given improper integral. We will first express this as a limit of definite integral. Is using its definition now by definition, this is equal to the limit as B approaches infinity of the integral from one to be of D X over x rays to the fourth power. Now this can be rewritten into the limit as B approaches infinity of the integral from 12 B of x rays to negative for dx. And then you will use power rule for integral. And this will give us limit as B approaches infinity Of x rays. The negative for post one over negative four plus one evaluated from one to be which is equal to the limit. As B approaches infinity of negative one over three times x rays to the third power That's evaluated from 1 to be. Now evaluating when X S B we have limit as B approaches infinity of negative 1/30 times be raised to the third power minus negative. That we plus one over three times 1 race to the 3rd power. And that's equal to the limit. As the approaches infinity of negative 1/3 times to be Cube plus 1/3. And evaluating this limit, we get negative 1/3 times infinity Plus 1/3. And since this goes to infinity and one over infinity goes to zero, we have zero plus 1/3. That's equal to 1/3. And because this is finite, we say that the improper integral converges

So we're looking at the integral from 0 to 1 of the function one over X to the pipe. Our so evaluating this integral with respect to X over the pi X to the pie power is going to be kind of difficult to find that anti derivative. So we're going to use our comparison, Farum. So we're going to set f of X, the one that were using equal to one over X to the pie and then our comparison we're going to set g of X as a more simple version of this So one over X. So that's just gonna be one over X to the one. So between zero and one, we're going to have f of X is greater. Thank G of X I'm If we drew Drew this on a graph, we would see that if we draw f of X in blue, we have f of X going like that And if we draw G of X in blue in green, we have g of X going like so. So between this barrier at one and this barrier at zero f of X is going to be greater than G of X at all times. So we're going to have using our comparison theorem. We're going to assume if G of X is diversion, then f of X is also divergent. So we're going to take the integral of G of X and hope that it diverges. And if not, we're just going to have to choose a different function for G of X. So we're gonna have one over X to the one evaluated with respect to X. So that's going to be equal to natural log of X evaluated from 0 to 1, we can't actually have natural log of zero. So we're going to have to do a limit here so we'll have the limit as he approaches zero from above of natural law of zero of your natural log of X, evaluated from t tow one. So plugging in one and T, we're going to have natural log of one minus limit as T approaches zero of natural log of tea. So this is going to be equal to zero minus something approaching infinity. So this is going to diverge. So since G of X is divergent than F of X must also be divergent, so given by our theory. And so we're gonna have our integral from 0 to 1 of the X over X to the pie well over X to the pie is going to be divergent.

And this problem. We're gonna take a look at this improper in a girl. And this is improper because that lower limit zero and the function is not to find there. The anti derivative of this one is Theo, Exact same as the previous problem in which I used partial for actions. And so what we're gonna do is turn this into her limit right away. We have are approaching zero from the right side and are integral will be from our team won and through the use of partial fractions, the anti derivative of this particular one is 1/5 the natural log the apse value of X divided by two x plus five. And again, we evaluate this from art of one. So we had the limit as our goes to zero from the right of 1/5 the natural log of the absolutely of one divided by two times one plus five, which is seven. They were just subtract 1/5 the natural log of the absurd value of are divided by two R plus five. When we look at this first part of our limit, this is a numeric value That's perfectly fine in regard to whether or not this converges it. That's a coverage of value. However, the issue becomes this second part. We plug in zero for our, um and we analyze this natural law algorithm. So just this portion right here, Really, What we get is essentially zero divided by five, and that's really zero. And the natural log of zeros as we approach from the right side, is gonna be negative. Infinity. And so this particular improper integral is going to be divergent.


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