The discussion revolves around solving the integral \(\int 41(\sin^{6}(x))(\cos^{3}(x)) \, dx\), which falls under the subject area of trigonometric integrals. Participants explore various methods and substitutions to approach the problem.
The conversation includes multiple interpretations of the problem, with some participants expressing confusion and seeking guidance. Others provide hints and encouragement to work through the integral, indicating a collaborative atmosphere. While some participants have made progress, there is no explicit consensus on the final approach or solution.
Participants mention feeling lost due to the recent introduction of the topic in class and the lack of teacher support. There is also a note of urgency as some participants balance this problem with other coursework.
Dick said:You don't have to do anything complicated with it, cos(x)dx=d(sin(x)). Just substitute u=sin(x). There are rules for dealing with powers of sin's and cos's. They are particularly easy if one power is odd.
Dick said:Oh, come on. Just work it out and show us what you get. I'll guarantee someone will check it.
Mugen Prospec said:Ok I have other work to do this one has bee on my mind all night. Ill post it tomorrow when I am clear of thought
Mugen Prospec said:Thats what I was thinking. after using u substitution I was left with
(u)^6 (cos^2(x)) cos(x) du/cos(x)
So cos(x) canceled out.
Mugen Prospec said:Oh ok the identity. so now all turns into
u^7 (1-u^2) du
Do I do their antiderivative now? And then substitute sin(x) back in. I am not sure if you can do each of there AD since they are multiplying one another.
Mugen Prospec said:OK that what I thought I am trying to do this not without witting it down since I am in chemistry.
So we get (u^6)-(u^8)du
then (u^7)/7) - (u^9)/9)
41 (sin^7(x))/7) - (sin^9(x))/9)
is that it maybe?