everybody knows integral and derivative cancel
by I know
so i will first convince you guys that the answer for this right here is indeed
sin^-1(x)+c and then i will show you guys how you can come up with your own integral magic
just like this as well in fact i came up with this because i was teaching my precalculus class
so you get 1/sqrt(1^2- (2x)^2) is just minus 4 x squared and then
because we're doing this kind of questions where we have to compose a regular trig function and
then inside the inverse trig function and we have to write this as the algebraic expression first
well the usual way to do this is that look at the inside when we have an inverse trig function that
just represents an angle so you can say let theta equal to that and then we can just take
the original sine on both sides so this means we get sine theta being equal to x which is the
same as x over one and right here because we're talking about sine this means we have the opposite
over hypotenuse from a right triangle and let's just go ahead and draw a right triangle put on
the right angle here and then the theta here the opposite will be here and the hypotenuse will be
here and the adjacent is just going to be the square root of the hypotenuse square minus the
other side squared and then to get secant of that we can just look at this picture
and we do hypotenuse over adjacent so this right here is just the integral of
1 over that right 1 over square root of 1 minus x squared and of course if you would
like you can do trig substitution whatsoever but you know that the answer for this is
inverse sine of x and of course don't forget the plus c and with that we are done
so after this i was like hm wow that's so cool because this right here is actually just the
derivative of inverse x so i thought about this a little bit more in fact this turns out to be
a really fun differential equation so this is how you can come up with your own integral magic
as well if you ever want to show people you can do that and get the answer this is the idea we want
to come with two functions so we'll just say that f and g to be the function that we want so that
when we integrate f and then the input is g right here and we really want to have that kind
of effect right meaning that we can just cancel cancel so we just want to get g of x back and
of course let's now put on the plus c because we can just do that at the very end later on
so what does this really mean this is an integral equation but we don't like that we like the
differential equation better so don't worry let's just go ahead and differentiate both sides with
dt like so when you do that the derivative and the integral can legitimately cancel out and
then we'll just get f of g of x and that's being equal to we will have to differentiate g of x
and of course this right here it's not a notation that we like when we
want to solve a differential equation so i'm just going to say let y be equal to g of x
because this way we can see that we have f as a function of y being equal to d y d x
much better and from here we can solve this differential equation by separating the variable
i'm going to multiply dx on both sides and i will divide the f of y on both sides but i will put
this part first the y on the left hand side so we get 1 over f of y with d y and that's equal to dx
and how in the world can we solve this now this is the time that you are going to pick
some function first and then just kind of work this out if you want to make that happen what
you can do is you can just start by saying let f of x be secant of x so this part becomes 1 over
secant of y d y and that's equal to dx and of course look at this this is not bad to integrate
because 1 over secant is just cosine so this is actually just cosine of y and then d y
that's equal to dx and then we can integrate both sides and you see integral of cosine is
just going to be past this side and then we have the y and then this is equal to just x and here if
you like and put down the plus c but to make things more clean i will say make c equal to
0 and then you can just take the inverse sine of both sides you get y is equal to inverse sine of x
and remember this is the y which is g of x which is the input input of f and the f
is our secant function therefore you see that the integral of secant of this right here goes inside
and when you do this you can really just cancel this and that and you end up with
inverse sine of x and at the very end just show people that don't forget the plus is right here
let's try another one so in order for this to work out nicely you have to make sure that the
f that you select it's easy to integrate when it's on the bottom so that's the key
another one to have in mind is tangent because when we have one over tangent is
actually just cotangent and it's actually not bad to integrate so let f of x being equal to tangent
of x this we get 1 over tangent of y d y and this right here is just equal to dx 1 over tangent is
cotangent of y and then we have the d y and z is equal to dx and of course now you can integrate
both sides and if you would like you can look at this as cosine of y over sine of y and do u sub
and either way you end up with ln absolute value of sine of y inside like this and then of course
the right hand side is equal to x plus c and again to make this cleaner let's pick c to be zero
and of course you can solve for y take e to the power here and e to the power here and then do
the inverse sine of both sides you get y is equal to inverse sine of e to the x how crazy is this
now i will put this back to here and then back to that integral so ladies and gentlemen we will
get the integral of tangent of this right here is for the g of x which is inverse sine of e to the x
this question can really scare a lot of kelp 2 students
but they wouldn't know that all you have to do is and then we'll get the answer
inverse sine of e to the x and of course now don't forget the plus c that's it
so with this right here for you guys you guys can kind of test it out and let me know what function
that you came up with and what integral magic that you come up with i think this is super super cool
and again i came up with this just by i would say purely coincident or yeah by luck and i think
the more i teach the more ideas i have because i've been doing this for like almost 10 years
and yeah i still have new ideas that's what makes math fun that's what makes teaching
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