The discussion revolves around differentiating the function \( \cos^2 x \) and understanding the implications of integrating the resulting expression. The subject area includes calculus, specifically differentiation and integration of trigonometric functions.
Several participants confirm the correctness of the differentiation process. However, questions arise regarding the integration of the differentiated result and the relationship between the integrated function and the original function, indicating a productive exploration of the topic.
Participants note the potential for differences in results when integrating, highlighting the importance of understanding the relationship between functions and their derivatives. There is an emphasis on the constant difference between the integrated forms of the functions involved.
Yes, your work looks fine.basty said:Homework Statement
What is ##\frac{d}{dx}(cos^2 x)##?
Homework Equations
The Attempt at a Solution
u = cos x
##\frac{du}{dx} = -\sin x##
##\frac{d}{dx}(cos^2 x) = \frac{d}{du}(u^2) \ \frac{du}{dx}##
##= 2u \ \frac{du}{dx}##
##= 2 \cos x (-\sin x)##
##= -2 \cos x \sin x##
Is it correct?
Yes. You may simplify it further using ##2\sin(a)\cos(a) = \sin(2a)##.basty said:Is it correct?
Mark44 said:Yes, your work looks fine.
It's possible for ##\int f(x)## to be equal to ##\int g(x)dx##, even though f and g aren't the same. However, they can differ by most a constant. In your case ##\cos^2(x) = -\sin^2(x) + 1##. In other words, these two functions differ by 1.basty said:Why the result is different when I integrate -2 cos x sin x back?
##\int -2 \cos x \sin x \ dx##
u = sin x
##\frac{du}{dx} = \cos x##
##du = \cos x \ dx##
##\int -2 \cos x \sin x \ dx##
##= \int -2 \sin x (\cos x \ dx)##
##= \int -2u \ du##
##= -\frac{2}{1+1}u^{1+1} + c##
##= -\frac{2}{2} u^2 + c##
##= -u^2 + c##
##= -(\sin x)^2 + c##
##= - \sin^2 x + c##