It WOULD be, but what we are really doing is using the definitions via the sides of a triangle... which, as you point out, are always positive. But don't forget that x is inside a square root (which is always positive itself), and when we do a back-substitution it still will be. It (almost) never happens that a situation arises where it would matter, so just like if our final answer had a ln|x| in it but we knew x was always positive, we'd write it ln x and get rid of the absolute value bars, we ignore the |sec x| and just cal it sec x because we MUST be in a region where that square root will always be positive anyway. I imagine it DOES get tricky if we are doing complex analysis, but my last complex analysis class was 30 years ago, and I don't remember if/how we handled trig sub for those problems... if we even did any.
Very nice! It can be solved also by IBP (x)' sqrt(1+x^2). Another nice substitution for this integral x=i sinu and you can use the identity i arcsin(-ix)=arcsinhx.
Let y : U -> R such that y is continuously differentiable everywhere, and such that D(y)(x)^2 - 1 = x^2 everywhere. This is equivalent to D(y)(x) = (-1)^m•sqrt(x^2 + 1) everywhere, with m = 0 or m = 1. Let g[m](x) = (-1)^m•sqrt(x^2 + 1) everywhere. Now, D(y)(x) = g[m](x) everywhere, with m = 0 or m = 1. Here, we can apply the Riemann integral on [0, t] to both sides of the equation, and by the fundamental theorem of calculus, the Riemann integral of D(y) is equal to y(t) - y(0). What remains is for us to find the Riemann integral g[m]. Notice that sqrt(x^2 + 1) = sqrt(sinh(arsinh(x))^2 + 1) = sqrt(cosh(arsinh(x))^2) = |cosh(arsinh(x))| = cosh(arsinh(x)) = cosh(arsinh(x))^2•1/sqrt(x^2 + 1) everywhere. This is important, because if h(x) = 1/sqrt(x^2 + 1) everywhere, then D(arsinh) = h. Also, notice that if j(x) = x/2 + sinh(2•x)/4 = x/2 + sinh(x)•cosh(x)/2 everywhere, then D(j)(x) = 1/2 + cosh(x)^2/2 + sinh(x)^2/2 = 1/2 + cosh(x)^2 - cosh(x)^2/2 + sinh(x)^2/2 = 1/2 + cosh(x)^2 - 1/2 = cosh(x)^2. As such, we can write that g[m] = (-1)^m•(D(j)°arsinh)•D(arsinh) = (-1)^m•D(j°arsinh) = D((-1)^m•(j°arsinh)), so the Riemann integral must be equal to (-1)^m•j(arsinh(t)) - (-1)^m•j(arsinh(0)). Thus, we have y(t) - y(0) = (-1)^m•(j(arsinh(t)) - j(arsinh(0))) = (-1)^m•j(arsinh(t)) everywhere. j(arsinh(t)) = arsinh(t)/2 + t/2•sqrt(t^2 + 1), hence y(t) - y(0) = (-1)^m/2•(arsinh(t) + t•sqrt(t^2 + 1)) everywhere.
Fantastic solution! I was just wondering, what if we bring x^2 to the LHS and take 1 to the RHS. Then we get a form: a^2 - b^2 = 1 Can't we then solve it like a Diophantine equation? It will be easy although the answers will be different of course, but is it wrong? Kindly explain it to me. Thank you.
Is that t substitution the same as tan(theta/2) substitution? When I see √(1+x^2) my inclination would be to go straight in with tan(theta) but I see someone has shown how that gives us sec^3(theta) which is a nuisance to integrate.
But my teacher always told me dy/dx is a symbol as a whole, not a damn division 😭😭😭 I do not know anything about calculus but I feel like that's incorrect. The integral has a dx at the end to make it clear that you are integrating with respect to x. And the integral with respect to x of the derivative with respect to x is the function. That's why on the other side there appears a "dx" because you integrated both sides with respect of x. Not because you multiplied both sides by dx. Am I correct? Or not? I'm just in highschool so I could be wrong
You are indeed correct, dy/dx is a symbol as a whole, but in this case you can treat it as division and it will not matter and it helps solving this equation easier.
I give you the third way. Integration "in parts". Int = ∫√(x^2+1)dx = | u=√(x^2+1) ⇒ du=x/√(x^2+1) , dv=dx ⇒v=x | = x*√(x^2+1)- ∫x^2dx/√(x^2+1) =x√(x^2+1)- Int +∫dx/√(x^2+1). | the last integral is considered tabular all over the world - "long logarithm"| 2* Int= x√(x^2+1)+ ln(√(x^2+1) +x)+C, Int= (1/2) *x√(x^2+1)+ (1/2)*ln[√(x^2+1) +x]+C, the sign of the absolute value of the logarithm argument does not need to be set. Positivity is fulfilled .)
This is an elementary problem. First thing it is of rank 1 unlike the Bessel differential equation for example . And 2nd thing the differential y' is solvable as a function of x . And after all we are still talking about ordinary differential equations !! ..... So it can be integrated directly .
nice contribution for this era people
That first substitution was filthy
The best substitution for the integral is x = sinh(t). Then you get integral of cosh(t)^2 where cosh(t)^2 = (1/2)(1+cosh(2t)) etc.
Yes, for sure. It falls out in half a dozen lines when you do that.
really.. you´re right..
The tan approach is amazing .. so we can replace the "ln" at the end with arc sinh(x)
In this kind of integrals is usually clever to use t=sinhx as cosh^2-sinh^2=1
I used the transformation x=tan(u) and obtained a shorter resolution because I obtained the integral of sec(u)^3 which is less complicated
Thank you for making more calculus and trigonometry videos!
you can directly use byparts in the question
10:30 Isn't the square root of (sec x)^2 absolute value of (sec x)?
It WOULD be, but what we are really doing is using the definitions via the sides of a triangle... which, as you point out, are always positive. But don't forget that x is inside a square root (which is always positive itself), and when we do a back-substitution it still will be. It (almost) never happens that a situation arises where it would matter, so just like if our final answer had a ln|x| in it but we knew x was always positive, we'd write it ln x and get rid of the absolute value bars, we ignore the |sec x| and just cal it sec x because we MUST be in a region where that square root will always be positive anyway.
I imagine it DOES get tricky if we are doing complex analysis, but my last complex analysis class was 30 years ago, and I don't remember if/how we handled trig sub for those problems... if we even did any.
Did all this 45 years ago, worked in process engineering and never used it. Advanced Stats would have been more of a help.
I don't like stats. No offense
Very nice! It can be solved also by IBP (x)' sqrt(1+x^2). Another nice substitution for this integral x=i sinu and you can use the identity i arcsin(-ix)=arcsinhx.
Great video!
More DEs would be great
When you replace ln|t|, why did you use t = sqrt(x^2+1)+x? shouldn't it be sqrt(x^2+1)-x?
Yes, there (9:10) should be either +ln(sqrt(x^2+1)+x) or -ln(sqrt(x^2+1)-x), because t^(-1) = sqrt(x^2+1)+x. He confused t and t^(-1).
Там действительно ошибка
Nice as well as educational.
👍✌️👍
WOW! Integral dy is really equal to y!
Darn… it’s been so long I don’t remember how to do any of these… I just use a spreadsheet now…
😲😄
👏👏👏👏👏👏👏
Awesome ♥️
Thanks 🤗
Long time, no see! 😊
please solve this is integral ln(cscx+cscx cotx) dx
its not an elementary integral.
Trig substitution is the best choice here
Can't we directly use integration by parts
y=(1/2)(xsqrt(1+x^2)+arcshx)+c
👍
Let y : U -> R such that y is continuously differentiable everywhere, and such that D(y)(x)^2 - 1 = x^2 everywhere. This is equivalent to D(y)(x) = (-1)^m•sqrt(x^2 + 1) everywhere, with m = 0 or m = 1. Let g[m](x) = (-1)^m•sqrt(x^2 + 1) everywhere. Now, D(y)(x) = g[m](x) everywhere, with m = 0 or m = 1. Here, we can apply the Riemann integral on [0, t] to both sides of the equation, and by the fundamental theorem of calculus, the Riemann integral of D(y) is equal to y(t) - y(0). What remains is for us to find the Riemann integral g[m].
Notice that sqrt(x^2 + 1) = sqrt(sinh(arsinh(x))^2 + 1) = sqrt(cosh(arsinh(x))^2) = |cosh(arsinh(x))| = cosh(arsinh(x)) = cosh(arsinh(x))^2•1/sqrt(x^2 + 1) everywhere. This is important, because if h(x) = 1/sqrt(x^2 + 1) everywhere, then D(arsinh) = h. Also, notice that if j(x) = x/2 + sinh(2•x)/4 = x/2 + sinh(x)•cosh(x)/2 everywhere, then D(j)(x) = 1/2 + cosh(x)^2/2 + sinh(x)^2/2 = 1/2 + cosh(x)^2 - cosh(x)^2/2 + sinh(x)^2/2 = 1/2 + cosh(x)^2 - 1/2 = cosh(x)^2. As such, we can write that g[m] = (-1)^m•(D(j)°arsinh)•D(arsinh) = (-1)^m•D(j°arsinh) = D((-1)^m•(j°arsinh)), so the Riemann integral must be equal to (-1)^m•j(arsinh(t)) - (-1)^m•j(arsinh(0)). Thus, we have y(t) - y(0) = (-1)^m•(j(arsinh(t)) - j(arsinh(0))) = (-1)^m•j(arsinh(t)) everywhere. j(arsinh(t)) = arsinh(t)/2 + t/2•sqrt(t^2 + 1), hence y(t) - y(0) = (-1)^m/2•(arsinh(t) + t•sqrt(t^2 + 1)) everywhere.
Fantastic solution! I was just wondering, what if we bring x^2 to the LHS and take 1 to the RHS. Then we get a form:
a^2 - b^2 = 1
Can't we then solve it like a Diophantine equation? It will be easy although the answers will be different of course, but is it wrong? Kindly explain it to me. Thank you.
what the
@@awindwaker4130 thanks. I realised it later. 😊👍
I started it wrong because I misread the square of the derivative as the second derivative. I wouldn't mind seeing the solution to that, btw.
=> Int (X^2•dx-dy)/2
Is that t substitution the same as tan(theta/2) substitution?
When I see √(1+x^2) my inclination would be to go straight in with tan(theta) but I see someone has shown how that gives us sec^3(theta) which is a nuisance to integrate.
Let J = integral{ [sec(x)]^3} dx = integral{sec(x) [sec(x)]^2} dx .
Let u=sec(x) & dv=[sec(x)]^2. So, du=sec(x)tan(x)dx & v=tan(x).
J = sec(x)tan(x) - integral{sec(x) [tan(x)]^2} dx
J = sec(x)tan(x) - integral{sec(x) ([sec(x)]^2 - 1)} dx
J = sec(x)tan(x) - integral{[sec(x)]^3} dx + integral{sec(x)} dx
J = sec(x)tan(x) - J + ln |sec(x) + tan(x)| + C
Solving for J:
J = (1/2) [ tan(x)sec(x) - J + ln | tan(x) + sec(x) | ] + C
Then back-sub to finish.
In the last line of equation (right after "Solving for J"), there obviously shouldn't be a J term on the RHS. :-(
@@timeonly1401 Yes because it has gone to the left 👍
I just cheated by looking at the integral table!
But my teacher always told me dy/dx is a symbol as a whole, not a damn division 😭😭😭
I do not know anything about calculus but I feel like that's incorrect.
The integral has a dx at the end to make it clear that you are integrating with respect to x.
And the integral with respect to x of the derivative with respect to x is the function.
That's why on the other side there appears a "dx" because you integrated both sides with respect of x. Not because you multiplied both sides by dx.
Am I correct? Or not? I'm just in highschool so I could be wrong
You are indeed correct, dy/dx is a symbol as a whole, but in this case you can treat it as division and it will not matter and it helps solving this equation easier.
I give you the third way. Integration "in parts".
Int = ∫√(x^2+1)dx = | u=√(x^2+1) ⇒ du=x/√(x^2+1) , dv=dx ⇒v=x |
= x*√(x^2+1)- ∫x^2dx/√(x^2+1) =x√(x^2+1)- Int +∫dx/√(x^2+1).
| the last integral is considered tabular all over the world - "long logarithm"|
2* Int= x√(x^2+1)+ ln(√(x^2+1) +x)+C,
Int= (1/2) *x√(x^2+1)+ (1/2)*ln[√(x^2+1) +x]+C,
the sign of the absolute value of the logarithm argument does not need to be set. Positivity is fulfilled .)
sub x with sinh(u) for easier approach
This is a standard problem in advanced level mathematics at high school. 😋😋😋😋😋😋
This is not very fast question
This is an elementary problem. First thing it is of rank 1 unlike the Bessel differential equation for example . And 2nd thing the differential y' is solvable as a function of x . And after all we are still talking about ordinary differential equations !! ..... So it can be integrated directly .
Podría hablar un poco más lento por favor
t or not to t
y=c±[x√(1+x²)+ln|x+√(1+x²)|]/2
👍