I think that u-sub of u=cos^-1(x) is actually a *much* easier and more direct approach, IMO. If you use Euler’s formula to treat the resulting sin(u) as the imaginary part of e^iu, then it becomes a simple integral of e^(1+i)u. This was straightforward enough that only took me a minute to work out the same final answer in my head (I was feeling too lazy to go fetch some paper, lol).
Yeah, but this one doesn't need Complex integrals, which aren't usually introduced during first two courses of analysis, I suppose. It's Just way more elementary, so almost any1 who just learned IBP can derive this or understand it basing only on knowledge, not believing some unseen theorems.
PoweredDragon but you didn’t notice that the substitution itself leads to another solution without complex integral. So are you admitting you are stupid enough that you cant figure it out or you didn’t read at all?
@@yuklungleung620 no, I didn't even touch the integral, I Just answered one exact comment and did the step-by-step in my head. That sort of integrals are not my concern any more at that point, so I'm not just sitting and solving it in 100 ways. Calling sb stupid just because you are not and you think that when you sit down and try sth other than the fact he's pointing out and he didn't look at... It's simply immature, so I suppose I'm not even talking to a person who understands calculus as sth more than just calculations.
This really doesn't suprise me if you think of cosine's exponential expression then arccosine is basically a logarithm of some stuff very similar to the logarithmic formula for the inverse hyperbolic functions. So taking e^arccosine is actually not too bad.
It makes sense that this would be integratable since in some sense exp(x) and cos(x) are the "same" function. Think cos(x)=½(exp(ix)+exp(−ix)) and exp(ix)=cos(x)+i*sin(x), so this is like f(a*f⁻¹(x)) but slightly different. This is actually exp(cos⁻¹(x))=cos(cos⁻¹(x)/i)+i*sin(cos⁻¹(x)/i), and if you look at it like that, it definitely seems plausible to integrate.
Hey,Dr peyam I think u substitution is better to use here let x=cosu dx=-sinu du now our original integral became ∫ e^u (-sinu) du By using integration by parts and substituting u=arccosx we get 1/2e^arccosx(x-√(1-x^2)) love from India
I am personally not very surprised it exists. If you think about it, e× is sort of a quasi trig function (because of Euler's formula), so its not unreasonable to expect that e^(inverse trig) will reduce to an expression we can integrate. Just like, for example, integral of tan(arcsin(x)) can be reduced to integral of x/√(1-x²) which is quite easy to integrate.
Wouldnt have been way easier to make the sub u=arccosx u=arccosx cosu=x -sinu du=dx we now have integral of e^u(-sinu) du which is very easy. Just substitute u in terms of x and use inverse trig identities to evaluate.
@@jonasdaverio9369 Guess I should have made myself more clearer. David Schmitz is right. Solve the integral in terms of u and then after solving the integral just substitute arccos x wherever u is and use trig identities and inverse trig identities to simplify
I remember there's a formula for the integral of the inverse of a function. Calling f^-1(x)= exp(arccos(x)), f(x)=cos(log(x)). I guess f is easier to see that we can integrate c:
Hey, problem suggester here. The “mess” you end up with by doing u-sub actually isn’t too bad, tho the steps you would use afterwards are essentially the same (keep doing IBP until you get the starting integral on both sides). But doing it that way would miss out on the beautiful simplification he showed in this video where the radical expression suddenly went away.
I would try to calculate it by parts twice, without substitution First integration D I e^arccos(x) 1 -e^arccos(x)/sqrt(1-x^2) x Second integration D I e^arccos(x) x/sqrt(1-x^2) -e^arccos(x)/sqrt(1-x^2) -sqrt(1-x^2) We can also choose parts in other way 1 = sqrt(1-x^2)/sqrt(1-x^2) so D I sqrt(1-x^2) e^arccos(x)/sqrt(1-x^2)
When I got arcsin(x) or arccos(x) in the integral, I tend to use u substitution for them. In this case I use u = arccos(x) which leds to integration by parts, twice.
It is much easier by this method Let cos-¹x= t Then we have integral of e to the power of t . As per the formula we know that integral of e to the power of t = e to the power of t + C. Put t= cos-¹x Then integral of e to the power of cos-¹x = e to the power of cos-¹x+C
I actually looked at that and thought "one half of a sum times the function", because cos x is (1/2)(exp(ix)+exp(-ix)). Also, the sum should have a term with a positive exponent and a term with a negative exponent. I mean, I didn't actually work it out, but I wasn't at all surprised by the form of the answer.
The prefix arc- is present in the names of the inverse trig functions because they return the arc length of the angle (the angle really because the circle has an unit radius anyway) for a given sine value. Because the area between the graph till (cosh t, sinh t) and x=y is just half of the input, the prefix used for them is ar- , not arc- , denoting the area. I wonder if there is any such analogy with the logarithm function and what would be the appropriate prefix to use for it? 🤔
@@angelmendez-rivera351 Yes, I was referring to the inverse hyperbolic trig functions but I think what I had written was wrong. I have edited my comment now, is it correct?
I used a substitutin x=cos(t), then arccos(x)=t, dx=-sin(t) dt, and the function under integral becomes: - e^t sin(t) dt, which is easily solved with the help of Euler formula. Of course, I have the same answer.
Doesn't seem that hard with u=cos^-1 (x). U get integral of -sinu e^u du. This can be easily integrated by parts twice to get back the negative of the original integral and u can just divide by 2 for the answer (not forgetting to substitute back into the x world and +c)
Can't we use the substitution t = exp(arcos(x)) which we can turn then into cos(ln(t)) = x? Then we differentiate to get dx and afterwards we can integrate t*(- sin(ln(t))/t) dt = - sin(ln(t))dt, which gives us the solution.
Yes you can do it. But here you would integrate sin(ln(t)) by parts. It is the same procedure as integrating exp(arccos(x)) by parts. The steps are similar. You get integral = some terms - integral, therefore you get 1/2*t* (sin(ln(t)) - cos(ln(t)) ) and then be careful with the resub. I would say to avoid a mess it is a good idea to integrate exp(arccos(x)) as it was shown.
@@drpeyam I just searched for it on his channel, and yes, he did it 4 years ago--here's the link: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-_MrNpNGaRdc.html
