oh my god. thank you so much. you went through every step, and even without voice over explaining it your work is just so clear and easy to follow. god bless you. you explained this wayyy better than my professor.
Thank you, Leela! I am very happy that it was useful for you! When I created this channel I decided to not add voice in order to let you think the reason of every step and to help you to do so I have to go through every step with the maximum of details possible. Of course, if the viewer doesn't understand something, it can be asked in a comment and I will try to answer the best I can! Thank you for your comment and welcome to my channel! 💪💪
Hi! It is solved almost the same way we did in the video: Integral of cos^3(2x)/sqrt(sin(2x)) dx = Substitution: u = 2x du = 2 dx ==> du/2 = dx = Integral of cos^3(u)/sqrt(sin(u)) du/2 = = (1/2)*Integral of cos^3(u)/sqrt(sin(u)) du = = (1/2)* integralsforyou.com/integrals?v=CoktCL4aZwQ = = (1/2)*[ 2sqrt(sin(u)) - (2/5)*sqrt(sin^5(x)) ] = = sqrt(sin(u)) - (1/5)*sqrt(sin^5(x)) + C Hope it helped! 💪
Hi! The result can be different if its derivative is the same... For example, in this case we can say that the answer is (1/5)sqrt(sin(x))(9 + cos(2x)) + C , but it is still the same answer: 2sqrt(sin(x)) - (2/5)sqrt(sin^5(x)) = = 2sqrt(sin(x)) - (2/5)sqrt(sin(x))sqrt(sin^4(x)) = = 2sqrt(sin(x)) - (2/5)sqrt(sin(x))sin^2(x) = = (1/5)sqrt(sin(x))[10 - 2sin^2(x)]= = (1/5)sqrt(sin(x))[9 + 1 - 2sin^2(x)]= = (1/5)sqrt(sin(x))[9 + sin^2(x) + cos^2(x) - 2sin^2(x)]= = (1/5)sqrt(sin(x))[9 + cos^2(x) - sin^2(x)]= = (1/5)sqrt(sin(x))[9 + cos(2x)] I have used: sin^2(x) + cos^2(x) = 1 cos^2(x) - sin^2(x) = cos(2x) So the question is: which is the result of your anthology?
Hi Md Tausif Raza! Here we have to do Weierstrass substitution t=tan(x/2) ==> th-cam.com/video/kmfMikTEIxM/w-d-xo.html t=tan(x/2) dx = 2/(1+t^2) dt sin(x) = 2t/(1+t^2) cos(x) = (1-t^2)/(1+t^2) Integral of sin^2(x)/(1+cos(x))^2 dx = = Integral of [(2t/(1+t^2))^2]/(1+(1-t^2)/(1+t^2))^2 2/(1+t^2) dt = = Integral of [(2t/(1+t^2))^2]/((1+t^2)/(1+t^2)+(1-t^2)/(1+t^2))^2 2/(1+t^2) dt = = Integral of [(2t/(1+t^2))^2]/((1+t^2 + 1-t^2)/(1+t^2))^2 2/(1+t^2) dt = = Integral of [(2t/(1+t^2))^2]/(2/(1+t^2))^2 2/(1+t^2) dt = = Integral of 2t^2/(1+t^2) dt = = 2*Integral of t^2/(1+t^2) dt = = 2*Integral of (t^2+1-1)/(1+t^2) dt = = 2*[ Integral of (t^2+1)/(1+t^2) dt - Integral of 1/(1+t^2) dt ] = = 2*[ Integral of 1 dt - Integral of 1/(1+t^2) dt ] = = 2*[ t - arctan(t) ] = = 2*[ tan(x/2) - arctan(tan(x/2)) ] = = 2*[ tan(x/2) - x/2 ] = = 2*tan(x/2) - x + C :-D
Hi, shibani chowdhury! Here it is: Integral of cos^4(x)/sin(x) dx = = Integral of cos^4(x)/sin(x) sin(x)/sin(x) dx = = Integral of cos^4(x)/sin^2(x) sin(x) dx = = Integral of cos^4(x)/(1-cos^2(x)) sin(x) dx = Substitution: u = cos(x) du = -sin(x) dx ==> -du = sin(x)dx = Integral of u^4/(1-u^2) (-du) = = Integral of u^4/(u^2 - 1) du = Polynomial division: D=dq+r u^4 = (u^2-1)*(u^2+1) + 1 u^4/(u^2-1) = (u^2-1)*(u^2+1)/(u^2-1) + 1/(u^2-1) = u^2 + 1 + 1/(u^2-1) = Integral of ( u^2 + 1 + 1/(u^2-1) ) du = = Integral of u^2 du + Integral of du + Integral of 1/(u^2-1) du = = u^3/3 + u + th-cam.com/video/X7OpiWaaArY/w-d-xo.html = = u^3/3 + u + (1/2)ln|u-1| - (1/2)ln|u+1| = = cos^3(x)/3 + cos(x) + (1/2)ln|cos(x)-1| - (1/2)ln|cos(x)+1| + C ;-D
+Sheikh Mahmud Hi! There isn't audio, maybe one day I will do it but by the moment the goal is to have the maximum number of videos with the solution. Thanks for comment!
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oh my god. thank you so much. you went through every step, and even without voice over explaining it your work is just so clear and easy to follow. god bless you. you explained this wayyy better than my professor.
Thank you, Leela! I am very happy that it was useful for you! When I created this channel I decided to not add voice in order to let you think the reason of every step and to help you to do so I have to go through every step with the maximum of details possible. Of course, if the viewer doesn't understand something, it can be asked in a comment and I will try to answer the best I can! Thank you for your comment and welcome to my channel! 💪💪
Thanks mate...this needed me on 2024... now
My pleasure! ❤
Thanks for your videos, they're really helpful
Thanks for your comment! ;-D
and what would be the result of this integral help me great sir please
Cos^3 2x/sqrt sin 2x dx
Hi! It is solved almost the same way we did in the video:
Integral of cos^3(2x)/sqrt(sin(2x)) dx =
Substitution:
u = 2x
du = 2 dx ==> du/2 = dx
= Integral of cos^3(u)/sqrt(sin(u)) du/2 =
= (1/2)*Integral of cos^3(u)/sqrt(sin(u)) du =
= (1/2)* integralsforyou.com/integrals?v=CoktCL4aZwQ =
= (1/2)*[ 2sqrt(sin(u)) - (2/5)*sqrt(sin^5(x)) ] =
= sqrt(sin(u)) - (1/5)*sqrt(sin^5(x)) + C
Hope it helped! 💪
Integration one upon under root cos^3(x).sin(x+a)=??
Sir apke es video se laga ki meraobile ka sound hi kharab ho gaya hai ❤❤
What will be the answer if we have
integrate (cos^3 x)/(sqrt(sin x)) dx from pi/3 to pi/2
Hi, Hmod KSA! Here you have the asnwer:
Integral of cos^3(x)/sqrt(sin(x)) dx from pi/3 to pi/2 =
= 2sqrt(sin(x)) - (2/5)sqrt(sin^5(x)) from pi/3 to pi/2 =
= [2sqrt(sin(pi/2)) - (2/5)sqrt(sin^5(pi/2))] - [2sqrt(sin(pi/3)) - (2/5)sqrt(sin^5(pi/3))] =
= [2sqrt(1) - (2/5)sqrt(1^5)] - [2sqrt( sqrt(3)/2 ) - (2/5)sqrt( sqrt(3)/2 )^5 ] =
= [2 - (2/5)] - [ sqrt(2)(3^(1/4)) - (2/5)( 3*3^(1/4)/4*sqrt(2) ) ] =
= 8/5 - (3^(1/4))(1/sqrt(2))[ 2 - (2/5)*(3/4) ] =
= 8/5 - (3^(1/4))(1/sqrt(2))[ 2 - 3/10 ] =
= 8/5 - (3^(1/4))(1/sqrt(2))(17/10) =
= 8/5 - (3^(1/4)/2^(3/2))(17/5) =
= 0.01797...
I have an anthology where the result is different, is the result of the anthology a mistake or what is it? Thank you
Hi! The result can be different if its derivative is the same... For example, in this case we can say that the answer is (1/5)sqrt(sin(x))(9 + cos(2x)) + C , but it is still the same answer:
2sqrt(sin(x)) - (2/5)sqrt(sin^5(x)) =
= 2sqrt(sin(x)) - (2/5)sqrt(sin(x))sqrt(sin^4(x)) =
= 2sqrt(sin(x)) - (2/5)sqrt(sin(x))sin^2(x) =
= (1/5)sqrt(sin(x))[10 - 2sin^2(x)]=
= (1/5)sqrt(sin(x))[9 + 1 - 2sin^2(x)]=
= (1/5)sqrt(sin(x))[9 + sin^2(x) + cos^2(x) - 2sin^2(x)]=
= (1/5)sqrt(sin(x))[9 + cos^2(x) - sin^2(x)]=
= (1/5)sqrt(sin(x))[9 + cos(2x)]
I have used:
sin^2(x) + cos^2(x) = 1
cos^2(x) - sin^2(x) = cos(2x)
So the question is: which is the result of your anthology?
sin^2x/(1+cosx)^2 integrate by substitution method. please solve it sir.
Hi Md Tausif Raza! Here we have to do Weierstrass substitution t=tan(x/2) ==> th-cam.com/video/kmfMikTEIxM/w-d-xo.html
t=tan(x/2)
dx = 2/(1+t^2) dt
sin(x) = 2t/(1+t^2)
cos(x) = (1-t^2)/(1+t^2)
Integral of sin^2(x)/(1+cos(x))^2 dx =
= Integral of [(2t/(1+t^2))^2]/(1+(1-t^2)/(1+t^2))^2 2/(1+t^2) dt =
= Integral of [(2t/(1+t^2))^2]/((1+t^2)/(1+t^2)+(1-t^2)/(1+t^2))^2 2/(1+t^2) dt =
= Integral of [(2t/(1+t^2))^2]/((1+t^2 + 1-t^2)/(1+t^2))^2 2/(1+t^2) dt =
= Integral of [(2t/(1+t^2))^2]/(2/(1+t^2))^2 2/(1+t^2) dt =
= Integral of 2t^2/(1+t^2) dt =
= 2*Integral of t^2/(1+t^2) dt =
= 2*Integral of (t^2+1-1)/(1+t^2) dt =
= 2*[ Integral of (t^2+1)/(1+t^2) dt - Integral of 1/(1+t^2) dt ] =
= 2*[ Integral of 1 dt - Integral of 1/(1+t^2) dt ] =
= 2*[ t - arctan(t) ] =
= 2*[ tan(x/2) - arctan(tan(x/2)) ] =
= 2*[ tan(x/2) - x/2 ] =
= 2*tan(x/2) - x + C
:-D
Easy quistion
Sir can u solve cos^4x/sinx integral...
Hi, shibani chowdhury! Here it is:
Integral of cos^4(x)/sin(x) dx =
= Integral of cos^4(x)/sin(x) sin(x)/sin(x) dx =
= Integral of cos^4(x)/sin^2(x) sin(x) dx =
= Integral of cos^4(x)/(1-cos^2(x)) sin(x) dx =
Substitution:
u = cos(x)
du = -sin(x) dx ==> -du = sin(x)dx
= Integral of u^4/(1-u^2) (-du) =
= Integral of u^4/(u^2 - 1) du =
Polynomial division: D=dq+r
u^4 = (u^2-1)*(u^2+1) + 1
u^4/(u^2-1) = (u^2-1)*(u^2+1)/(u^2-1) + 1/(u^2-1) = u^2 + 1 + 1/(u^2-1)
= Integral of ( u^2 + 1 + 1/(u^2-1) ) du =
= Integral of u^2 du + Integral of du + Integral of 1/(u^2-1) du =
= u^3/3 + u + th-cam.com/video/X7OpiWaaArY/w-d-xo.html =
= u^3/3 + u + (1/2)ln|u-1| - (1/2)ln|u+1| =
= cos^3(x)/3 + cos(x) + (1/2)ln|cos(x)-1| - (1/2)ln|cos(x)+1| + C
;-D
shouldn't U^2 / U^ -1/2 = U^5/2?
u^2/u^(-1/2) = u^(2-(-1/2)) = u^(2 + 1/2) = u^(5/2) but in this integral we have
u^2/u^( 1/2) = u^(2-( 1/2)) = u^(2 - 1/2) = u^(3/2)
I see it now, thanks.
You're welcome! :-D
Thanks
You're welcome!! ;-D
where's the audio???
+Sheikh Mahmud Hi! There isn't audio, maybe one day I will do it but by the moment the goal is to have the maximum number of videos with the solution. Thanks for comment!
Thanks
You're welcome! ;-D