Integral of sqrt((1-x)/(1+x))

here's how to integrate x/sqrt(1-x^2)

Integral of sqrt((1-x)/(1+x))
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[sr.tarsaimsingh]

We can also proceed with 1st method
Use Substitution X=cos(A)
Then, solve and we get solution as
=√(1-x^2) - arccos(X) + C
Try like this you will find new approach.

[ritsnay]

I was solving this question by 1st method and I got stuck so I just searched this question on Google and now I'm here nice explanation ...

[__lucky___]

I am from India your explanation is very well and your face expression very nice ☺️👍

[cobalt]

I am currently taking Multivariable Calculus (moving onto Linear Algebra second semester), am on the math team for my school, and have taken AMC, AIME, and ARML tests in the past with remarkable results. Yet when I look at all these integrals and see how all these functions you didn't think had anything to do with one another are all related and interconnected in such complicated ways...it still just completely blows my mind.

[kymosahbe]

I know this is may be old video lol but this really helped me with my Calculus, I'm currently taking Calculus 2 in college and I need all the help I can take lol
Thank you BlackpenRedpen!

[rohitsinha]

Did anyone notice how smoothly he shifts betn pens and uses them

[GSHAPIROY]

In (pre)algebra, we rationalize the denominator. In calculus, we rationalize the numerator.

[studyhardforyear]

You can also substitute (cos2theta) in place of x

[harveyarizala]

YOU DON'T KNOW HOW MUCH YOU SAVED MY GRADES. THANK YOU VERY MUCH!

[itdoesntmatter]

We can also use substitution substitute x=cos theta then we can integrate

[studyvibes]

Thanks very much sir. I'm facing the problem to solve this sum, so i searched this question on google and now I'm here good explanation. Again thanks sir😇😇

[Anonymous-kwls]

Alternative method..
just rationalise the given function.. The numerator root will be cancelled by square..
next put x = sint.. now by substitution you will get the answer..

[aravartomian]

Except there is a logical error because you are assuming that x does not equal to 1; otherwise you are multiplying by 0/0, which is undefined. Or you could let x=(y^2)-1 in which case the integrand reduces to 2(2-y^2)^1/2 after cancelation of dx=2ydy on the top and y on the bottom. The above integrand can be taken care of by trig substitution.

[festinalente]

Thanks for explanation sir. You helped me so much. Salutes from Spain.

[ritsnay]

We can also substitute cos2thita in place of x

[cemsentin]

I use x=cos2u and dx=-2sin2u*du transormation,

Sqrt[(1-cos2u)/(1+cos2u)] became,

=Sqrt[2(sinu)^2/2(cosu)^2]

=Sqrt[(sinu/cosu)^2]

=sinu/cosu

=tanu

Hence,

Int(Sqrt[(1-x)/(1+x)]*dx)

=Int[tanu*-2sin2u*du]

=Int[-4sinu*cosu*tanu*du]

=Int[-4(sinu)^2*du]

=Int[-4*(1-cos2u)/2*du]

=Int[(2cos2u-2)*du]

=sin2u-2u+C

After using x=cos2u, 2u=Arccosx an inverse transformation,

[futureworld]

Great!!!.GOOD LUCK for your future.☺️☺️☺️

[eahype]

You deserve my Professor's salary.

[premvirsingh]

My problem had a x multiplied in numerator which is actually making it tougher, this questions is so simple

[cbaadith]

How do u integrate if the numerator was root x-1 instead and denominator is still root 1+x