integrate-each-of-the-given-functions-int-fracsec-2-t-tan-t4sec-2-t-d-t-2

Question

Integrate each of the given functions.
$$\int \frac{\sec ^{2} t 	an t}{4+\sec ^{2} t} d t$$

Amy Jiang · Accepted Answer

Perkins looks up for letting you equal people in t squared to. And now if we take derivative with that, we get to t beachy and inter numerator. We have t g so itself for that. So we had a deal over two is equal to t achieve. Okay, so we get in a role. Um, do you over to an anti squared costs to that vehicle to you. Now let&#x27;s pull out our what happened. So we get do you over you and that&#x27;s people to the island of active value of you. Let&#x27;s see. And now let&#x27;s replace. Are you so yet? Um T squared, plus two and in custody.

Foster Wisusik · Answer

Okay, This question wants us to integrate the following function. So since we&#x27;re dealing with trig functions here, we should probably use our U substitution in our argument with that trick function. So let&#x27;s pick you equal to two tea. So this means that d u is just two times d t. But we don&#x27;t have to d t anywhere in our integral we only have one d t. So let&#x27;s divide both sides of that equation by two to get do you over two equals DT So now we can plug in and get that are integral is now just the integral c Can&#x27;t you tangent you times do you over too or 1/2 times the integral of C Can&#x27;t you tangent you? Do you and this anti derivative right here is just one of our basic ones for Trig It&#x27;s the anti derivative which just gives us He can&#x27;t because the derivative of seeking seeking tangent. So this is just 1/2 see, can&#x27;t you plus c or 1/2 c can&#x27;t of to t plus c when we substitute back in for you

Aishwarya Krishnakumar · Answer

problem. Nine were asked to find the definite in true. And to do this, we are going to first take out the constant. So in this case, the concept is too, so you can take that out and rewrite it, and then it makes it easier for us to just solve for the integral. So if you add one wide to over two, then it would be t three Hobbs times to over three minus. If you add one, it would be he squared over two. And this whole thing, since it&#x27;s a definite integral, would be evaluated from 0 to 4. So now all you have to do is plug in four and then minus the value you get when you plug in zero for tea. Well, let&#x27;s go ahead and do that. It would be 2/3 four since you&#x27;re bugging and four for the T values minus four squared over too. This minus 2/3 times zero to the three halves. Power minus zero squared over, too. And if you compute this, you will get to crimes negative. Eight over three, which equals negative. 16 over three

Xiaomeng Zhang · Answer

so we can say those one dressy close to two times into grow off the square Rules off T right now two times in Q grow off t and see what does one that echoes who true, comes to over three. He to the power offs trio were true in the Rangers here in full for those one grotty co two t Squire over to zero. Well, so that he comes to 4/3 through the power of history. That is a to minus zero. And for those one were how ah, just he Esquire he calls two of 16. See here we how circuit to over three minutes. Ah 16 that IHS 48/3. So on how 16/3 miners

Amy Jiang · Answer

Okay, so we&#x27;ll stop by letting you equal Equalling our exponents Power threats to t minus described and now are derivative. That&#x27;s equal to to punish to t beachy and number stats. We almost have it in the form be achieved times one minus t. So it&#x27;s factor out of you and then divide by two on both sides. So we get to you over two is equal to one minus t h e. Okay, so now let&#x27;s let one wanna see the t v do you over to, and then you have B to the power of you. It&#x27;s bring out our 1/2 and now let&#x27;s take her in. And girls, so we get 1/2 each other you put See an older place are, um Are you? So are you Is, uh, seeing that&#x27;s duty minus piece word and then plus seat

Problem

Integrate each of the given functions. $$\int \fr…

Problem 32 Easy Difficulty

Integrate each of the given functions.
$$\int \frac{\sec ^{2} t \tan t}{4+\sec ^{2} t} d t$$

Answer

$\ln \left|4+\sec ^{2} t\right|+C$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Kayleah T.

Caleb E.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$\ln \left|4+\sec ^{2} t\right|+C$

Basic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Kayleah T.

Caleb E.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

Anna Marie V.

Kayleah T.

Caleb E.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus