integrating and differentiating sin^2(x) and cos^2(x) (1 Viewer)

[helz_h]

Help. I simply can't do this. I have the wrong formula in my notes and would be very appreciative of someone correcting it!

Thanks a million!

 

[Trev]

d/dx * sin²x = 2cosx.sinx
d/dx * cos²x = -2sinx.cosx

Like you know how cos²x+sin²x=1
d/dx * [cos²x+sin²x] = d/dx * 1
2cosxsinx - 2sinxcosx = 0
0 = 0

 

[onebytwo]

for integration
make since cos²x=(1/2)(1+cos2x)...then integrate
do the same thing for sine

 

[m0ofin]

Use the function of a function rule after you take the square out.
y = sin²x
y = (sinx)²
y' = 2cosx.sinx

Same for cosx

 

[helz_h]

thanks for your help!

Good luck with your exams