Martyn Arthur said:
You guys are so patient. Given the following
xr^3(1-cos^2θ)cos^2θ
I need to make the substitution u = (1-cos^2θ) have I got it right?
Thanks
View attachment 340095
I can't make any sense of this of the attachments. They are the same image double-posted and are as follows:
$$xr^3(1 - \cos^2)(\cos^2)$$
What does this mean? Why is ##\theta## missing from the attachments?
What are you supposed to do with this?
The work you show also makes little sense, partly because some of it is cut off, partly because it's hard to decipher, and partly because much of the rest doesn't make sense.
At the top you have "Let ##u = (1 - \cos^2\theta)##" and then <something> u = ##\cos u## <something>
In the 2nd equation on the left side you have a squiggle -- is it f? Or is it an integral sign? On the right side the top of the fraction is cut off. Is where I have ? supposed to be d?
On the next line you have ##\frac d {d\theta} = -\cos^2 \theta##
This doesn't make sense for a couple of reasons:
1) On the left side you have a differentiation operator, but are missing whatever it is that you are differentiating? Should this have been ##\frac{du}{d\theta}##?
2)Assuming that you meant ##\frac{du}{d\theta}## on the left side of the second line, the right side is not the derivative of ##1 - \cos^2\theta##
If the goal is to find the derivative of ##1 - \cos^2 \theta##, your answer at the bottom is correct, but there is a lot of cruft and incorrect stuff along the way.
Note that ##1 - \cos^2 (\theta) = \sin^2(\theta)## so
##\frac d {d\theta}(1 - \cos^2 (\theta)) = \frac d {d\theta}(\sin^2(\theta)) = 2\sin(\theta) \cdot \frac d {d\theta} (\sin(\theta)) = 2\sin(\theta) \cdot \cos(\theta) = \sin(2\theta)##