the trigonometric substitution skip

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[Bodyknock]

Considering that trig functions are entangled with the Pythagorean Theorem, it makes sense that you can often use a Pythagorean type equation for the substitution (e.g. u ² = 1 - x ² is literally just the formula for the lengths of the sides of a right triangle of lengths u, x, and hypotenuse 1, and it’s also an easy way to prove that 1 = sin ²x + cos ²x as well. Likewise u ² = 1 + x ² is a triangle with a base length 1, height x, and hypotenuse u, making u linked to the tangent of the corresponding interior angle.)

So really in a way you’re just skipping the trig notation and going directly to playing with the corresponding squares of the lengths of the right triangle they represent on a unit circle.

[wmeisel]

Shouldn’t the second one end up as -1/3((1-x^2)/(x^2))^3/2+C

[bjornfeuerbacher]

Since I've learned about hyperbolic functions, I much prefer to use the substitution x = sinh(t) for integrals containing sqrt(1+x²), instead of x = tan(theta). Usually leads to an integral containing only exponential functions, and you don't need to use lots of tricky trig identities.

[byronwatkins]

At 8:30, you omitted the x^2 in the denominator when you substituted in for the final form.

[order]

I always understand these videos, but would never be able to come up with these substitutions on my own.

[MacHooolahan]

Yes! You've finally given voice to something that has troubled me for years here. I *could* call it "how many symbols do you really need"....

[zalut_sky]

it's just binomial differential if you think about it. good video, Michael!

[matematicagoiania]

This video has an excellent observation professor! But, I guess you got a mistake on the last integral result; probably, it should be or +C or (-1/3)[((x^(-2))-1)]^(3/2)+C, shouldn’t it?

[MinhTran-flqg]

for those looking for the original article, its doi is

[kenhill]

A lot of it is practice and trial and error. Michael obviously has years of working these problems so he can quickly recognize potential substitutions. However he oresents a single substitution, there may be several that would work. As well he's not working these problems on the fly. You can tell he's looking at notes so he's already tried several substitutions until he find one that works and then presents that.

It would be fun to see him produce a video on his process of how he picks problems and works them out (mostly showing all the stuff that didnt work - this is where the real Math happens 99 failures for 1 success but a lot of fun exploration along the way).

As for a substitution the big thing you are looking for is a continuous function whose range is the domain of what you are integrating. So your integral is on some set S you ideally want a continuous function from P->S then you can integrate on P. Obviously there are A LOT of subtleties i just hand waived over.

[marcvaneijmeren]

With the second integral you should state that x > 0. Otherwise you have a sign change when the 1/x is placed under the root.

[theupson]

none of the variations of either example plays very differently from if you just do the trig sub. the first sub for all intents and purposes IS the standard trig sub, the second one is effectively u=csc(theta); the second problem played by the book you factor the trig into csc and cot to prosecute the resulting trig integral.
i do not believe there is any advantage to be gained here.

[ianfowler]

It is interesting to see how his new substitutions are in some way "mimicking?" the trig identities. So for the first one, u^2 = 1 + x^2 when x = tan(t) then u becomes sec(t) and in the second one u^2 = 1/x^2 - 1, u becomes cot(t).

[julianamaths]

I wouldn't have done the first one with trigonometric anyway, I'd just have done u=1+x^2, which also turns out quite simple. But for the second one, u=1-x^2 does NOT turn out simple, so it is possible that I would have tried trigonometric for that one, but it turns into cos^2/sin^4, which is also very painful, so at that point I would have probably paused to look for a third way. I love this trick of pulling one of the x's inside of the square root that you are showing us here, I'll definitely keep that in mind for the future, but I am having trouble seeing how to generalize the kinds of situations where this trick applies? It feels to me that this was one very lucky example, if it was any other exponent there instead of 4, it wouldn't have turned out that nice. What am I missing?

[syphaxjuba]

merci sire excellent exercice, est il possible faire avec substitution hyperbolique

[matthewuzhere]

i would prefer using u=sqrt(1+x^2) in the first one. as you stated it’s essentially equivalent, but it would avoid the absolute value issue that you ran into with sqrt(u^2).

[dufflepod]

I liked that second integral. It came out so cleanly.

[jay_sensz]

There's one weird substitution trick that works on any integral: Let u = F(x) such that dF/dx = f(x).
1du = f(x)dx
so we have the integral over 1 du, which is clearly u + c
Substituting back, we get F(x) + c -- Easy! 😎

[johnvandenberg]

I’m a big fan of his videos, even though he often makes small calculation errors, like here again 😁

[cycklist]

Can you demonstrate an integral where you do a really non-obvious substitution. Like the met effect of three or something.