Hi! Since the integral of f'(x)cos(f(x)) dx = sin(f(x)) , where f'(x) is the derivative of f(x), in our case we have f(u) = 2u and its derivative is f'(u)=2. We need to multiply cos(2u) by 2 in order to have its derivative: Integral of cos(2u) du = = (1/2)Integral of 2*cos(2u) du = = (1/2)sin(2u) If you don't like this method, you can use substitution: Integral of cos(2x) dx = th-cam.com/video/GtXYyJqsHak/w-d-xo.html
hi ı have one question why we say x=cos(u) instead of x=sin(u) ?? I was solution with x=sin(u) and answer is =x^2/2*arccosx+1/4(arcsinx-xsqrt(1-x^2)+C These answers are same?
Hi, Ibrahim, how are you? Since: Derivative of arcsin(x) = 1/sqrt(1-x^2) Derivative of arccos(x) = -1/sqrt(1-x^2) Then both next expressions have the same derivative: (x^2/2)arccos(x) - (1/4)arccos(x) - x*sqrt(1-x^2) + C (x^2/2)arccos(x) + (1/4)arcsin(x) - x*sqrt(1-x^2) + C Then both answers are the same and they are the solution of the integral of x*arccos(x). In this video I chose x=cos(u) in order to have the solution in terms of arccos(x).
Hi jj ms! 1. cos(arccos(x))=x By definition, if cos(x)=y then x=arccos(y). So, x=arccos(y)=arccos(cos(x)). 2. sin(arccos(x))=sqrt(1-x^2) We can use the formula 1=sin^2(y)+cos^2(y) ==> sin(y)=sqrt(1-cos^2(y)). Let y=arccos(x): sin(arccos(x))=sqrt(1-cos^2(arccos(x)))=sqrt(1-x^2) Hope it helped!
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thank you so much this helped me a lot :) But could you write bigger next time if possible
I'll try! You can watch my last videos, they have more quality, let me know what you think 😉
@@IntegralsForYou i didn't realize this video was 4 years old lol. I will watch them. Thanks for the positive feedback :)
@Duru Thanks for your comments! 💪💪
at 3:49 where did the 2 infront of cos come from?
Hi! Since the integral of f'(x)cos(f(x)) dx = sin(f(x)) , where f'(x) is the derivative of f(x), in our case we have f(u) = 2u and its derivative is f'(u)=2. We need to multiply cos(2u) by 2 in order to have its derivative:
Integral of cos(2u) du =
= (1/2)Integral of 2*cos(2u) du =
= (1/2)sin(2u)
If you don't like this method, you can use substitution:
Integral of cos(2x) dx = th-cam.com/video/GtXYyJqsHak/w-d-xo.html
I wonder why sqrt(sin²(u))＝sinu instead of ±sinu
Hi! I found this article that explains it: math.stackexchange.com/questions/1118400/trig-substitution-why-can-we-ignore-the-absolute-value 😉
@@IntegralsForYou thanks a lot😘
@@wyl3638 My pleasure! ☺
hi ı have one question why we say x=cos(u) instead of x=sin(u) ??
I was solution with x=sin(u) and answer is =x^2/2*arccosx+1/4(arcsinx-xsqrt(1-x^2)+C
These answers are same?
Hi, Ibrahim, how are you?
Since:
Derivative of arcsin(x) = 1/sqrt(1-x^2)
Derivative of arccos(x) = -1/sqrt(1-x^2)
Then both next expressions have the same derivative:
(x^2/2)arccos(x) - (1/4)arccos(x) - x*sqrt(1-x^2) + C
(x^2/2)arccos(x) + (1/4)arcsin(x) - x*sqrt(1-x^2) + C
Then both answers are the same and they are the solution of the integral of x*arccos(x). In this video I chose x=cos(u) in order to have the solution in terms of arccos(x).
Nice ;)
:-D
why sin(arccos(x))=square(1-x2) and cos(arccos(x))=x?
Hi jj ms!
1. cos(arccos(x))=x
By definition, if cos(x)=y then x=arccos(y). So, x=arccos(y)=arccos(cos(x)).
2. sin(arccos(x))=sqrt(1-x^2)
We can use the formula 1=sin^2(y)+cos^2(y) ==> sin(y)=sqrt(1-cos^2(y)). Let y=arccos(x):
sin(arccos(x))=sqrt(1-cos^2(arccos(x)))=sqrt(1-x^2)
Hope it helped!
Integrals ForYou Thanks, you saved my life. Let's see if I pass the calculus exam next week
I wish you the best for your exam, and if you have more questions here I am! ;-D