Integral sqrt(tanx) from zero to pi/2 using phase shifting

Another cool integral!
In this video I've explained how to use phase shifts to make life easier while integrating trig functions. I hope you enjoy video and be sure to like and subscribe to help the channel grow.

[renesperb]

Nice trick. Another way is as follows : use the substitution tan x= t^2 . Then you get the integral of 2 t^2 /(1+t^4), which you can calculate using partial fractions, or more elegant, residues to get π/√2 .

[maalikserebryakov]

I learned more integration techniques and strategies from you than any book

[txikitofandango]

A lot of great development over the last 5 months

[Ni]

My favorite indefinite solution, you probably recognize it, is -
(1/√2)arcsin(sinx-cosx)
- + c

I never stopped to notice that when the bounds are complimentary the second term is canceled out. Pretty neat.

Thanks, this was a great solution, really enjoying your work. 👍

[BernhardElsner]

Another method consists of substitution t = tan(x) and then use a classical residue calculus.

[digxx]

Let I=Int(tan(x)^s, x=0..Pi/2)=Int(u^s/(1+u^2), u=0..infinity) for s€(-1, 1) and by the substitution u=tan(x). Consider the integral Int(u^s/(1+u^2), u=-infinity..infinity) with the cut on (-infinity, 0) and the path of integration above the cut. For s in the given range you can close the contour in the upper half plane and use the residue theorem to find the value 2Pi*i*i^s/(2i)=Pi*i^s. On the other hand Int(u^s/(1+u^2), u=-infinity..infinity) = Int(u^s/(1+u^2), u=-infinity..0) + Int(u^s/(1+u^2), u=0..infinity) = This gives I=Pi/2/cos(Pi/2*s).

[ChrT]

When substituting, you say twice "you do not need a constant of integration here".
I think it would be better to say, "It is entirely your choice what you are substituting here, so you can choose the constant to be zero or anything else".

[MathNotationsVids]

Outstanding videos - - Great explanations 🤩

[johnathan]

I can get more better with calculs.Thank you.🙂.Any book suggestions for 1st year?

[mokouf]

Very similar trick can be used for indefinite version of this integral too!

[krisbrandenberger]

@ 2:47 The integral in red at the bottom should have a dx, and not a dt.

[JayadrataBanerjee]

how did root(cott)dt become root(cotx)dx? i don't understand

[matteocilla]

can you try to solve int of tan(sqrt(x)) dx?? thx

[Asterisme]

Why not directly employ the king's property to immediately have:
I(√(tan(x));0;π/2)) =
I(√(tan(π/2-x));0;π/2))

[harryjega]

What if you multiply the two equations after substitution and the original equation.
You will get I^2 =int 1 dx from 0 pi/2
Which is I^2 =pi/2
And I = sqrt (pi/2)

[sandeepmehrotra]

Actually the formula integralfrom 0---> π/2 (tan{x})^1/n =[ π/2 ]÷[Cos{πn/2}] is more convenient

[cooldawg]

I dont understand this “difference is in NAME ONLY” comment. “T” was defined earlier as x + pi/2 so it doesnt seem they are just the same variable

[cooldawg]

You say “t” and “x” are the same, but didnt you explicitely DEFINE “x” as “pi/2” - u?

I dont understand this just saying “u” is “x” part