answer-the-given-equations-related-to-the-indicated-examples-of-this-section-in-example-1-how-must-2

Question

Answer the given equations related to the indicated examples of this section.
In Example 1, how must the integrand be changed in order to have
a result of $\sin ^{-1} x+C ?$

Rukhmani Jain · Accepted Answer

in discussion. It is told that we&#x27;re able to make some changes in example number request so that our integration of the function should be signing for sex. Okay so I&#x27;m going to make some changes and I&#x27;m going to write the integration of one upon square will talk one minor success square Bx. Okay now let&#x27;s start with this integration. I&#x27;m going to make a substitution here and that is I am going to replace access by scientific tire. Okay now we&#x27;ll see the creative is you can see here scientists derivative is cause data data. So carefully. I&#x27;m going to make this replacements letter did not integration by eye. So you can see that what we should get here after making these replacements one divided by one minus sine and square theater. And at the place of the X. We have to substitute cause theatre de theatre. Okay now further see what happens here. Trigonometry. Identity says that square root of one minus sign is worth it to is there and one minus sign is worth it. Again. The bitterness causes were theater and further we have caused to the theater. Okay, now square out. Of course where theater is simply I can say that one divided by cost data. And we have to multiply by cost it also we can cancel our cost you to so we get here one day theater and now we can integrate one with respect to theater and its integration should be close to theatre plus C. Okay, since we are solving an indefinite integration so I add added constant also. Now I&#x27;m going to re substitute here. Okay now I&#x27;m going to re substitute. You can see that we had taken Your acts of the question by scientific to hear. It means that data must be replaced by signing over six. Okay, so if you make the substitution, I get here signing what sex. Let&#x27;s see. Okay, this is the answer. And you can see that to get this answer. What tired do you see here that I can say here is integration of one upon square root one minus access where bX. Okay. Thank.

Geena Pullo · Answer

eso in this question, We&#x27;re going to be solving the indefinite integral using the method of you substitution. So we have the integral of the coast of X over the sign of X times the square root of the sign of X squared plus one de acts. So first thing going to do is just going to rewrite this as the integral of coast of X over sign of acts times the square root of sign X squared plus one squared. It&#x27;ll make sense more sense later. I just want to do that now. So we have it written that way. So the first thing I&#x27;m going to do now is set my sign equal to are you So you is equal to sign of Axe and from here we&#x27;re gonna take the derivative of both sides. So we&#x27;re going to have d u is equal to the coasts of X dx. And so from here we&#x27;re going to just divide both sides by our coast of X. So we&#x27;re gonna have d you over the coast of access equal to T X. And now we&#x27;re going to replace R sign of X with argue and RDX with R D you over coast of acts. So now we&#x27;re going to have the equation. So they were the integral of the coast of acts over you Times the square root of you squared plus one squared eu over the coasts of X and our coasts of axes coasts is of exes are going to school across out and we&#x27;re just going to be left with the integral of the G You over the u times the ah square root of you squared plus one squared. So from here, we&#x27;re going to be using the equation that states that the integral of D u over you times the square root of you squared. Uh, sorry. A squared plus U squared is equal to negative one over a times the Ellen or the natural log of a plus the square root of a squared plus u squared over you plus C. So because we have Ah, plus here plus is a really interchangeable. But just so we make a note are a is going to be equal to one. And are you, of course for now is equal to you so we can just plug in, um, to this equation from what we have on well, now have, um, solved, at least for the integral. So we&#x27;re going tohave Negative one, because are a is equal to one. So we&#x27;re going toe have just one times the l n of one plus the square root of one squared Plus you squared over you plus C. And so from here, all we&#x27;re going to do is plug in for are you and are you is equal to the coasts of Sorry about the coast. That&#x27;s the sign of X, right? Yes, you is equal to the sign of acts. So we&#x27;re just going to have the equation that this is this integral is equal to negative. L and I have one plus the square root of one cause one squared is equal to one plus sign of X squared over sign of axe plus C. And that is our answer

Julian Gerber · Answer

So here we have the integral of co sign X divided by sine x times square root of sine squared X plus one. So to do this, we&#x27;re going to start off with using a use substitution so we can say u equals sine X and therefore d U equals, um co sign X d x So we have the co sign X dx and the numerator so we can just replace that with d u perfect sine X just becomes you So we have you times the square root of you squared plus one Now to go from here, we&#x27;re going to use a trig substitution. So instead of you, we&#x27;re gonna set you now equal to the tangent of another variable Let&#x27;s say Z So do you equals C can&#x27;t squared z d Z And the reason we&#x27;re doing this is because ultimately we are going to want to use the, um, Trig identity. Tan squared axe plus one equals C can&#x27;t squared X. And if we&#x27;re able to use that identity than the intro becomes a lot easier. So let&#x27;s do another substitution. So in the numerator instead of d u, we have c can&#x27;t squared Z D c over tangent of Z times the square weird of tan squared Z plus one. Now we see tansy tan squared Z plus one can use this identities We get the square root of C can&#x27;t squared Z which of course is just see can&#x27;t see So now what we have If we simplify as I c can&#x27;t squared z DZ and the numerator then we have tan the times this see Kanzi and the denominator Now from there one of disease can or one of this he can ZEEZ can cancel out. So we&#x27;re left with just, um the integral of c can&#x27;t z over tangent Z jeezy. Now, if we simplify that c can&#x27;t is one over co sign, right and tangent is sign over co sign. So if we simplify that just using, you know, multiply, um, both the numerator and dominator by co sign over sign. Then what we end up getting is, um the what we end up getting is the integral of, um, one over sign gets one over co sign divided by, um, sign over. Co sign equals one over sign, which is the same thing as co. C can&#x27;t. So now we have finally the integral of Cosi cans. And we can use just a, um, integral table to find that the integral of coast he can&#x27;t is just negative. Ln Cosi can&#x27;t z plus co tangent Z plus Of course, the constant of integration. Now that we have technically are integral, all we have to do is change these ease back into exes. So let&#x27;s free up some room, Teoh. Change it all back into terms of X. We&#x27;re going to use the, um, equations. We initially substituted four. So remember, our substitution is in the very beginning, we had u equals sine X and then later, in order for a trick substitution, we had u equals tangent Z. So if we want to relate, um, Z toe acts, we get that tangent Z equals sine X. Now, if we want to solve for, um Z in terms of X, um, what we can do is just take the inverse tangent of both sides so we get Z equals inverse tangent of sine X. Now, we can use this substitution to substitute, um, that into our answer here to get it back into terms of X. So we have negative Ln absolute value of Cosi can&#x27;t now instead of Z, we have in verse tan sine X plus co tangent of inverse tan of sine X plus the constant of integration and that is our final answer.

Foster Wisusik · Answer

Okay, This question wants us to find the integral of this function. First using a series of substitution tze and then just doing one big substitution. So first, let&#x27;s just pick you equal to X minus one Because that&#x27;s the argument of our trick function and simply enough We just get d X equals do you? So now we have the integral of square root one minus signs square Do you your sight Plus rather times sign you co signed you, do you? But now let&#x27;s pick Z equal to sign of you So we get even you is equal to co sign you, do you which we do have So now we just get the integral of square root one plus fi squared times see TV because signed you again is just equal to be so Lastly what we can d&#x27;oh is have w equal one plus fees squared So DW is just to v d me or dw over to is equal to VT So now, for our last substitution, we just get the integral of squared of w times dw over too. And this is just 1/2 times w sir. Sorry. Sums are inside derivative which is 2/3 W to the three halves plus C, and this is equal to once. We simplify 1/3 times W to the three halves, and we said W was equal to one plus fees squared, and we said that V was equal to sign of you. And then we said that you was equal to X minus on, and this is our final answer. And again we could have achieved the same thing. By starting out, we&#x27;ll do another integral here. And this time, let&#x27;s just make one big substitution. So let&#x27;s just say that omega is equal to one plus signs squared of X minus one. So this means the D omega is equal to two sine X minus one co sign X minus one by the chain role. The ex or de omega over to is equal to Sign X minus one co sign X minus one. So now we can integrate and get 1/2 times the integral of square root of omega de omega or 1/2 times 2/3 omega to the three halves. Let&#x27;s see for plugging back in, we get 1/3 and then omega was one plus sine squared of X minus one plus C. And our anti derivatives do indeed match

Patrick Vaughn · Answer

consider the enrol of the X over X minus X squared. And if we use the substitution, you equal Teoh exponents 1/2 do you unequal dx and we would get do you over you plus 1/2 my ass, you plus 1/2 square. So then that would get us you plus 1/2 minus you squared. Plus you plus 1/4. That would be do you over So the you know the one power terms they&#x27;re gonna cancel, We&#x27;ll get 1/4 minus you squared and then using the formula from 8.1 that gets us in verse sign of you over 1/2 plus some constant. Now, if we plug you back in as expense, 1/2 will get sign in verse, uh, two x minus one Wilson constant. So we&#x27;ll call this Formula One. Now, suppose we take the same in that role p x overs for X minus X squared. And instead of using you equal to X minus 1/2 we used you equal to root out. So then do you is 1/2 x dx where we can write that as to you dio people to dx. This is also the X is useful. So when we use that substitution, we&#x27;ll have to you, do you over square of you squared minus you to the fourth. So that&#x27;s two D U Over. I swear of one minus. You squared. This is going to be too Denver Sign you. It&#x27;s a constant. Finally, we can write to sine inverse of root X plus some constant We call that formula to now, If we take formula to finance Formula One, we can find that to sign inverse proof acts fine US sign members of two X minus one is gonna be equal to some constant. Now, the reason why I can say this is because work these were both formed from the integral of D X over root X minus X squared. So we have two of acts is equal to this, and one of X is equal to this. So like it, they can only vary by a constant because formula to prime of X is equal to one over root X minus X squared and Formula One prime of X is equal to one over root X minus X squared. All right, so he&#x27;s gonna call this formula three. So for bringing three down. We have to sign inverse fruit X minus sine inverse of two X minus one is equal to some constant. This is true for any value of X. Let&#x27;s just begin at X equals zero we have to sign and various of zero. My signed in verse of negative one is equal to some constant Sinan versus 00 sign in verse. The negative one is negative. Pi over two. So we get zero plus pi over two. A sequel to see This proves to sign in various through X minus sine inverse of two x minus one is equal Teoh by over two.

Ernest Castorena · Answer

thank you of this problem. They want us to show that don&#x27;t kiss Nancy. Derivative can be equal to two different things. That&#x27;s a different but miracle. This is just the derivative, uh, coastline in verse. All right, And then we need to add in arbitrary. But we can use our integration role to pull the negative out and recall that this is just the derivative of our coasting. Because I never That&#x27;s equal to the above. Sign it. Thanks the negative in front of it, plus him for some arbitrary constant to avoid confusion that this arbitrary constants that one. I&#x27;ll call this C two and I&#x27;ll call this one c one and they want to say explain why it does not imply. But co sign Inrix of X is equal to negative sign and worse, right? Actually, when we&#x27;re careful like this, making sure they&#x27;re different constants, it&#x27;s already clear they may not be equal. Now let&#x27;s see industry that these 200 rolls of the same So it is true that these two equations should be equal and we could have them. Let&#x27;s subtract them from each other so well, get zero on that side is equal to co sign inverse of a plus. He won, um, and subtract negative sine inverse effect track. See? So we have Ah, bring it to the other side C to minus C. One is equal to co sign inverse of X plus. Sign it recently, pretty me and one equation on constant that would make thes satisfying is pi over two. And you can think about how you might get that well, thinking about their how their original functions regulated, aren&#x27;t they just ships of each other? You have saying of, like, an angle? Well, consider coastline, uh, by over to attract that angle as a function. But this is already good enough of an equation to relate them awesome.

Problem

Answer the given equations related to the indicat…

Problem 1 Easy Difficulty

Answer the given equations related to the indicated examples of this section.
In Example 1, how must the integrand be changed in order to have
a result of $\sin ^{-1} x+C ?$

Answer

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Basic Technical Mathematics with Calculus