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Integral of sqrt(x^2+1), integral of sqrt(x^2-1), integral of sqrt(1-x^2)

bprp calculus basics

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In this calculus tutorial, we will do the classic trigonometric substitution problems, namely, the integral of sqrt(x^2+1), integral of sqrt(x^2-1), and integral of sqrt(1-x^2). These are some must-know examples for your Calculus 2 class! Best wishes to you in your class!

Calculus 2 Integration techniques: Playlist

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20 Comments

@bprpcalcbasics Creator 5 years ago Pinned

Watch next: What integration technique do we use?
https://youtu.be/HiXfAayQ_8o

@mrbobma-z3w 7 months ago

You only person that solve that at that time when i am search it for my exam

@PixelPiewe 4 months ago

preparing for cuet exam in india - thanks for helping out man

@grassytramtracks 1 year ago (edited)

It comes out much nicer if you do x=sinhu for the first one, and x=coshu for the 2nd

@rogersmith-h2f 3 years ago

Gracias hermano me resolviste el examen eres un artista de las matemáticas.

@jhouck1969 1 month ago

It helps to visualize the identity tan²θ + 1 = sec²θ by drawing a unit circle, then extending the tangent to the x-axis and forming a right triangle where the radius (1) and tanθ are the legs, and secθ is the hypotenuse.

@SaidiAmri-h8g 9 months ago

I like the way of teaching

@MeraimBouchabchoub 1 year ago

شكرا من المغرب❤❤❤

@mathunt1130 5 years ago

I'd be tempted to do a hyperbolic sub in the first integral. It comes out easier.

@fixtir947 2 years ago

thanks, man I seen a lot of videos and I liked how you explained how to come back to the x world the best. I was very confused.

@user-wu8yq1rb9t 5 years ago

I have a recommendation (or request).
Please make a video about the *Wallis Product*.
Thank you

@akashbhatti5141 4 years ago

Thank you sir

@ajshots748 3 months ago

Do int(sqrt(x^2-1)) and int(sqrt(1-x^2)) have some sort of imaginary relationship? I say because they are same if you multiply one of them by i you get the other.

@ChrisRossaroDidatticaDigitale 3 years ago

In the first one I'd use this substitution: x=ch(t); in this way, by using ch^2(t)-sh^2(t)=1 and dx=sh(t)dt, one gets int{sh^2(t)dt}; thus, integrating by part =1/2*{sh[settsh(ch(x))]-settch(x)}+c. Through the hyperbolic goniometric definitions you get the very same result. Thank for the video.

@nostalgiez3429 3 years ago

Fastest way to do integals like that is by integration by parts just let initital function to be u ( u = sqrt(x^2 + 1)) for example, you will get answer pretty quick

@Shahan1980 3 years ago

I can't remember even one problem from higher schoo; or university where we used sec or cosec? I mean I know what it is, but didn't even write the formulas involving those two. sin and cos was always enough? though, ok, fractions might appear, but, from the other point of view trig functions are all about fractions, just meticulously hidden