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

[@IntegralsForYou]

👋 Follow @integralsforyou for a daily integral 😉 📸 instagram.com/integralsforyou/ 𝐈𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝𝐬 𝐩𝐥𝐚𝐲𝐥𝐢𝐬𝐭 ► Integration by parts kzbin.info/aero/PLpfQkODxXi4-GdH-W7YvTuKmK_mFNxW_h ► Integration by substitution kzbin.info/aero/PLpfQkODxXi4-7Nc5OlXc0zs81dgwnQQc4 ► Integration by trig substitution kzbin.info/aero/PLpfQkODxXi49OUGvTetsTW61kUs6wHnzT ► Integration by Weierstrass substitution kzbin.info/aero/PLpfQkODxXi4-8kbKt63rs1xwo6e3yAage ► Integration by partial fraction decomposition kzbin.info/aero/PLpfQkODxXi4-9Ts0IMGxzI5ssWNBT9aXJ 𝐅𝐨𝐥𝐥𝐨𝐰 𝐈𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬 𝐅𝐨𝐫𝐘𝐨𝐮 ▶️ KZbin kzbin.info 📸 Instagram instagram.com/integralsforyou/ 👍 Facebook facebook.com/IntegralsForYou 𝐃𝐨𝐧𝐚𝐭𝐞 🙋‍♂️ Patreon www.patreon.com/integralsforyou

[@redpilled7209]

let sin(t)=sqrt(x), cos(t)=sqrt(1-x), t=arcsin(sqrt(x)) 2sin(t)cos(t)dt=dx integral of 2cox^2(t) = t + 1/2 sin(2t)+c = arcsin(sqrt(x))+sqrt(x)sqrt(1-x)+c

[@IntegralsForYou]

💪💪

[@profesordanielalvarez3498]

Este canal vale oro en polvo! :D

[@IntegralsForYou]

En lingotes también me vale jeje gracias Daniel!!

[@smashliek5086]

At 1:00 you can use the formula of iint( sqrt(a^2 - x^2)dx)

[@IntegralsForYou]

Hi! If your teacher allows you to use it, then yes! 😉

[@pepitolacaca7844]

and what would happen if you do sqrt((x-1)/x) is it the same? and what would happen if need to do a definite integral in de interval of -2 to -1 lets say

[@IntegralsForYou]

Hi! Here you have the solution: Integral of sqrt((x-1)/x) dx = = Integral of sqrt(x-1)/sqrt(x) dx = Substitution: u = sqrt(x-1) ==> u^2 = x-1 ==> u^2+1 =x du = 1/2sqrt(x-1) dx = 1/2u dx ==> 2u du = dx = Integral of u/sqrt(u^2+1) 2u du = = 2*Integral of u^2/sqrt(u^2+1) du = = 2* kzbin.info/www/bejne/mp28ZZuIjJ2nl6s = = 2*[ (1/2)u*sqrt(u^2+1) - (1/2)ln|u+sqrt(u^2+1)| ] = = u*sqrt(u^2+1) - ln|u+sqrt(u^2+1)| = = sqrt(x-1)*sqrt(x) - ln|sqrt(x-1) + sqrt(x)| + C The definite integral between -2 and -1 cannot be calculated because the function is defined for x>=1 ;-) Let's calculate it between 1 and 2: Integral of sqrt((x-1)/x) dx from 1 to 2 = = [ sqrt(2-1)*sqrt(2) - ln|sqrt(2-1) + sqrt(2)| ] - [ sqrt(1-1)*sqrt(1) - ln|sqrt(1-1) + sqrt(1)| ] = = [ sqrt(1)*sqrt(2) - ln|sqrt(1) + sqrt(2)| ] - [ sqrt(0)*sqrt(1) - ln|sqrt(0) + sqrt(1)| ] = = [ sqrt(2) - ln|1+sqrt(2)| ] - [ - ln|1| ] = = [ sqrt(2) - ln|1+sqrt(2)| ] - [ - 0 ] = = sqrt(2) - ln|1+sqrt(2)| = = 0.5328...

[@shrutigupta9319]

Too good

[@IntegralsForYou]

Thanks! 😊

[@plslovetxt2607]

Hello, thank you for this video! But I wonder how to solve sqrt(x+1) / sqrt(x) ? I’m trying so hard to solve this by following ur ways but I’m stuck as i reached the sin cos part :(

[@IntegralsForYou]

Hi! Here you have the solution! Integral of sqrt(x+1)/sqrt(x) dx = Substitution: u = sqrt(x) ==> u^2 = x du = 1/2sqrt(x) dx = 1/2u dx ==> 2u du = dx = Integral of sqrt(u^2+1)/u 2u du = = 2*Integral of sqrt(u^2+1) du = = 2* kzbin.info/www/bejne/bKTWe5aKhreghLc = = 2*(1/2)[ u*sqrt(u^2+1) + ln|sqrt(1+u^2) + u| ] = = u*sqrt(u^2+1) + ln|sqrt(1+u^2) + u| = = sqrt(x)*sqrt(x+1) + ln|sqrt(1+x) + x| + C Hope it helped! ;-D

[@sidjdykakjskdnejdif2891]

How can you make u = sin(t) what about if u bigger than 1 or lower than -1. what t values give those u variables.

[@IntegralsForYou]

sqrt(1-x) ==> 1-x>0 ==> 1>x ==> x x>0 Then we have sqrt(1-x)/sqrt(x) is defined for x in (0,1). Since u=sqrt(x) and u=sin(t), then we have sqrt(x) = sin(t) in (0,1). Then the values we can take for t are the values that make sin(t) be in (0,1) 😉

[@sidjdykakjskdnejdif2891]

@@IntegralsForYou Oh I did not see that sorry. Thank you for your respond.

[@sidjdykakjskdnejdif2891]

I subscribed and I will continue to watch your hand 😂.

[@IntegralsForYou]

@sidjdykak jskdnejdif My hand is very happy to read your comment! 😊😊

[@sidjdykakjskdnejdif2891]

😂😂

[@bhanupratap8947]

Thanku much

[@IntegralsForYou]

😊

[@BomberKing]

Thank you

[@IntegralsForYou]

My pleasure! 😊

[@bryanginting513]

Is it ok to make u-substitution with numerator??because what i got is different to what you got when i make u-substitution with numerator

[@IntegralsForYou]

Hi! Yes, we can do u=sqrt(1-x) and we should have the same derivative. Let's take a look: Integral of sqrt(1-x)/sqrt(x) dx = Substitution: u = sqrt(1-x) ==> u^2 = 1-x ==> x = 1-u^2 ==> sqrt(x) = sqrt(1-u^2) du = -1/2sqrt(1-x) dx = -1/2u dx ==> -2u du = dx = Integral of u/sqrt(1-u^2) (-2u)du = = -2*Integral of u^2/sqrt(1-u^2) du = = -2* kzbin.info/www/bejne/aZrbhYtqYtWZjZI = = -2*[(1/2)arcsin(u) - (1/2)*u*sqrt(1-u^2)] = = -arcsin(u) + u*sqrt(1-u^2) = = -arcsin(sqrt(1-x)) + sqrt(1-x)*sqrt(x) + C The solution in the video is arcsin(sqrt(x)) + sqrt(1-x)*sqrt(x) + C. Let's check the derivative of -arcsin(sqrt(1-x)) is the same than the derivative of arcsin(x): Derivative of -arcsin(sqrt(1-x)) = = - [1/sqrt(1-(sqrt(1-x))^2)]*[1/2sqrt(1-x)]*(-1) = = [1/sqrt(1-(1-x))]*[1/2sqrt(1-x)] = = [1/sqrt(1-1+x)]*[1/2sqrt(1-x)] = = [1/sqrt(x)]*[1/2sqrt(1-x)] = = 1/[2sqrt(x)sqrt(1-x)] Derivative of arcsin(sqrt(x)) = = [1/sqrt(1-(sqrt(x))^2)]*[1/2sqrt(x)] = = [1/sqrt(1-x)]*[1/2sqrt(x)] = = 1/[2sqrt(x)sqrt(1-x)] Since both derivatives are equal, then we can say that both solutions are correct 😉

[@bryanginting513]

@@IntegralsForYou Well this is new for,since only thing that i know,sometimes we integrate,we can get different constant when we use other methods or substitution,but now there are 2 different possible form of functions that we can get,so it's confusing at the first,but now i know we can derive of that different function to make sure that is same.Thanks for the rigorous explanation😀

[@IntegralsForYou]

@@bryanginting513 My pleasure! 😊 Do you know my website? integralsforyou.com Enjoy it! 💪

[@Kun6mu]

Hi, I find Your videos very useful. Could you please help me integrate 1*dx/(1+x^4)^(1/4)?

[@RamSingh-bd8fc]

Plc don't cover whole page with your hand.......its a request

[@IntegralsForYou]

I'll try to do it better the next time! 💪

[@aayanfaiz3086]

Thank s m

[@IntegralsForYou]

My pleasure! 😊

[@tanseerahmad6340]

What's that arcsin?

[@IntegralsForYou]

Hi! It means arcsine, the inverse function of sine, sometimes also written as sin^-1. Hope it helped! 😊

[@dariak-j7q]

Hey. you have a similar example where the numerator and denominator are reversed

[@IntegralsForYou]

Hi! I have some examples in this playlist: kzbin.info/aero/PLpfQkODxXi49kD2n2dCoSGRsscd5TME_Z If you don't find what you are looking for, write it here in a comment and I'll try to solve it in a comment too 😉

[@IntegralsForYou]

Integral of sqrt(x)/sqrt(1-x) dx = = Integral sqrt(x)*(1/sqrt(1-x))dx = Substitution: u = sqrt(1-x) ==> u^2 = 1 - x ==> x = 1 - u^2 du = -1/2sqrt(1-x) dx ==> -2 du = 1/sqrt(1-x) dx = Integral of sqrt(1-u^2) (-2du) = = -2Integral of sqrt(1-u^2) du = = -2 kzbin.info/www/bejne/eJupeaRpfs2UbLs = = -2(1/2)( arcsin(u) + u*sqrt(1-u^2) ) = = -arcsin(u) - u*sqrt(1-u^2) = = -arcsin(sqrt(1-x)) - sqrt(1-x)sqrt(x) + C 😉

[@dipalisutradhar4008]

Sir please sum intregration of e^x(1+x^2)dx/(1+x)^2

[@ernestschoenmakers8181]

Well this integral is done by u-sub and integration by parts, here we go: Let x+1=u then du=dx and x=u-1, the integral becomes: int of e^(u-1)*(1+(u-1)^2)/u^2*du= int of e^(u-1)*(u^2-2u+2)/u^2*du= int of e^(u-1)*(1-2/u+2/u^2)*du, now we split the integral into 3 parts: int of e^(u-1)*du - int of 2e^(u-1)/u*du + int of 2e^(u-1)/u^2*du= e^(u-1) - int of 2d(e^(u-1))/u - int of 2e^(u-1)d(1/u), now we apply integration by parts: e^(u-1) - 2e^(u-1)/u + int of 2e^(u-1)d(1/u) - int of 2e^(u-1)d(1/u), the 2 integrals cancel so we'll get: I = e^(x) - 2e^(x)/(x+1) + C = e^(x)*(1 - 2/(x+1)) + C = e^(x)*((x-1)/(x+1)) + C.

[@norareyes3622]

wow you are so smart can´t relate

[@IntegralsForYou]

Thanks, Nora! 😉

[@bhargavikotturu2659]

Sound Add cheyyu raa

[@teddyfatimdiallo617]

But your hand is hiding it..

[@IntegralsForYou]

But not always...

[@bahadurverma6316]

Goonga thik se bata

[@dhruvyadav7138]

Thank you

[@IntegralsForYou]

My pleasure! 😎