Integral of cos^5(x)/sqrt(sin(x)) (substitution)

[@IntegralsForYou]

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[@reemrahmaoui6158]

Hi! but isnt derivative of cosx = -sinx? instead of sinx being the u?

[@IntegralsForYou]

Hi! We do u=sin(x) because we have all the expression in terms of sin(x) except for one cos(x) which is next to the "dx". Since du=cos(x)dx we can substitute cos(x)dx by "du". If we want to do u=cos(x) then du=-sin(x)dx which is -du=sin(x)dx. However, we need all the expression in terms of cos(x) and going this way may be more complicated than the first one: Integral of cos^5(x)/sqrt(sin(x)) dx = = Integral of cos^5(x)/sqrt(sin(x)) sin(x)/sin(x) dx = = Integral of cos^5(x)/sin(x)*sqrt(sin(x)) sin(x)dx = = Integral of cos^5(x)/sqrt(sin^3(x)) sin(x)dx = Substitution: u = cos(x) ==> u^2 = cos^2(x) ==> 1-u^2 = 1-cos^2(x) ==> 1-u^2 = sin^2(x) ==> sqrt(1-u^2) = sin(x) ==> (1-u^2)^(1/2) = sin(x) ==> [(1-u^2)^(1/2)]^3 = sin^3(x) ==> (1-u^2)^(3/2) = sin^3(x) du = -sin(x)dx ==> -du = sin(x)dx = Integral of u^5/sqrt((1-u^2)^(3/2)) (-du) = = - Integral of u^5/(1-u^2)^(3/4) du = = - Integral of u^4/(1-u^2)^(3/4) u*du = = - Integral of (u^2)^2/(1-u^2)^(3/4) u*du = Substitution: t = 1-u^2 ==> u^2 = 1-t dt = -2u*du ==> (-1/2)dt = u*du = - Integral of (1-t)^2/t^(3/4) (-1/2)dt = = (1/2)*Integral of (1-2t+t^2)/t^(3/4) dt = = (1/2)*Integral of [1/t^(3/4) - 2t/t^(3/4) + t^2/t^(3/4)] dt = = (1/2)*Integral of t^(-3/4) dt - Integral of t^(1/4) dt + (1/2)*Integral of t^(5/4) dt = = (1/2)*t^(1/4)/(1/4) - t^(5/4)/(5/4) + (1/2)*t^(9/4)/(9/4) = = (1/2)(4/1)*t^(1/4) - (4/5)*t^(5/4) + (1/2)(4/9)*t^(9/4) = = 2*t^(1/4) - (4/5)*t^(5/4) + (2/9)*t^(9/4) = =[t^(1/4)]*[ 2 - (4/5)*t + (2/9)*t^2 ] = =[(1-u^2)^(1/4)]*[ 2 - (4/5)*(1-u^2) + (2/9)*(1-u^2)^2 ] = =[(1-cos^2(x))^(1/4)]*[ 2 - (4/5)*(1-cos^2(x)) + (2/9)*(1-cos^2(x))^2 ] = =[(sin^2(x))^(1/4)]*[ (2 - (4/5)*sin^2(x) + (2/9)*(sin^2(x))^2 ] = =[(sin(x))^(1/2)]*[ 2 - (4/5)*sin^2(x) + (2/9)*sin^4(x) ] = =sqrt(sin(x))*[ 2 - (4/5)*sin^2(x) + (2/9)*sin^4(x) ] + C

[@bravek9881]

Thank you so much but i cant understans where did we get the 4/5 at 5:08..help please

[@IntegralsForYou]

Hi Brave K! -2 u^(5/2)/(5/2) = -2/(5/2) u^(5/2) = = -2 (2/5)u^(5/2) = = -(2*2/5)u^(5/2) = = -(4/5)u^(5/2)

[@bravek9881]

THANK YOU SO MUCH now it is all clear!♦

[@IntegralsForYou]

You're welcome! :-D

[@juniorschmitt]

if cos^2, who is the better substituition?

[@IntegralsForYou]

Hi! The integral of cos^2(x)/sqrt(sin(x)) is a non-elementary integral, it cannot be expressed in terms of finite standard functions...

[@juniorschmitt]

Thank you very much for responding, now I am aware. My teacher had made a typing error in the list of exercises.

[@IntegralsForYou]

@@juniorschmitt Well, I think it is non-elementary... if I am wrong please let me know hehe

[@shreyanshuyadav4505]

Thank u so much💯👍

[@IntegralsForYou]

You're welcome! 😊😊

[@kenedyq.f9917]

eres lo maximo

[@IntegralsForYou]

jeje gracias, Madrilista 10! :-D

[@fatiherim5859]

i follow you in turkey. Im waiting others integrals examples. thanks

[@IntegralsForYou]

Hi furkan Erim! Next week I will upload new videos! Thanks for asking, and let me know if there is an example that you want me to do! :D

[@fatiherim5859]

Integrals ForYou help me for integral cot^3x

[@IntegralsForYou]

Hey! Take a look at this video! kzbin.info/www/bejne/Z6WTiadueax9jJI In your case: cot^3(x) = cos^3(x)/sin^3(x) = ((1-sin^2(x))/sin^3(x)) cos(x) And you do substitution: u = sin(x), du = cos(x)dx Integral of ((1-sin^2(x))/sin^3(x)) cos(x) dx = Integral of ((1-u^2)/u^3) du =....

[@IntegralsForYou]

Integral of cot^3(x): kzbin.info/www/bejne/eZe8e36EidKin7M

[@bestnocture]

So satisfying to watch you integrate!

[@dipakray339]

Thank you boss😃😃😃

[@IntegralsForYou]

You're welcome! ;-D

[@СнежныйБарс-х8ь]

288

[@IntegralsForYou]

Nice to see you again! ♥️

[@СнежныйБарс-х8ь]

@@IntegralsForYou ✋😉🫖☕🍋🐅🐆🐅🐆🐅🐆

[@cedrickvillanueva8272]

I didn't understand at 5:53 untill 5:55

[@IntegralsForYou]

I simplified the square root: sqrt(u^4) = u^(4/2) = u^2

[@tiuk23]

sin(x)=u^2, cos(x)*dx=2u*du is better substitution in my opinion.

[@nidhisingh5925]

Bhai hath hatiye ga,tab na kuch dikhega