Trig Integration: Integrating (sinx)^3 * cosx with Respect to x

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Homework Help Overview

The problem involves integrating the expression (sinx)^3 * cosx with respect to x, which falls under the topic of trigonometric integration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of u substitution as a method for integration, with some questioning the specifics of how to apply it. There are attempts to clarify the process of identifying u and du in the context of the given integral.

Discussion Status

Several participants have provided guidance on using u substitution, suggesting different approaches based on the powers of sine and cosine. There is an ongoing exploration of how to apply these concepts, with some participants expressing uncertainty about their understanding of u substitution.

Contextual Notes

Some participants indicate a lack of familiarity with u substitution, and there is mention of the problem being off-syllabus for one participant, which may affect their ability to engage with the discussion fully.

QueenFisher
integrate with respect to x: (sinx)^3 * cosx

i have no idea where to start, can anyone help me? I've looked at differentials of other trig functions but i can't see any that would help :mad:
 
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what do you mean?
 
you want to use a u substitution. Find a value in your expression to be u, and find another one to be du.

~Lyuokdea
 
QueenFisher said:
what do you mean?

How would you integrate [tex]\int (x^2+7)^3 \cdot 2x \,dx[/tex] ?
 
Use U sub. as indicated before. Let your U = sin(x) . Work it from there
 
QueenFisher said:
integrate with respect to x: (sinx)^3 * cosx

i have no idea where to start, can anyone help me? I've looked at differentials of other trig functions but i can't see any that would help :mad:
If your sine function is raised to an odd power, it's commonly to let u = cos x, and work from there.
If your cosine function is raised to an odd power, it's commonly to let u = sin x, and work from there.
If both are raised to an odd power, then you can either let u = sin x, or u = cos x.
Note that, you should sometimes need to use the Pythagorean identity: sin2x + cos2x = 1, to solve your problem.
I'll give you an example:
-----------
Example:
[tex]\int \cos x \sin ^ 2 x dx[/tex]
cos x is raised to the power 1, hence it's an odd power, let u = sin x.
u = sin x => du = cos x dx, right? Substitute that into your integral, we have:
[tex]\int u ^ 2 du = \frac{u ^ 3}{3} + C[/tex]
Change u back to x, gives:
[tex]\int \cos x \sin ^ 2 x dx = \frac{\sin ^ 3 x}{3} + C[/tex]
Can you go from here? :)
 
i've never done anything like that before and i don't quite understand it
 
Let me see if I can help you. :smile:

~Kitty
 
  • #10
Let me see if I have this correct, you have:

sine of x cubed times cosine of x right?

~Kitty
 
  • #11
Please don't take this as attacking. Have you been exposed to u substitution? I'm assuming you have.

~Kitty
 
  • #12
My process was about the same as VietDao. I'm sorry.

~Kitty
 
  • #13
Too many posts... :\
Dude, if you didn't take this in class, read off this site for help. :}
 
  • #14
misskitty said:
Please don't take this as attacking. Have you been exposed to u substitution? I'm assuming you have.

~Kitty

unfortunately not. but it turns out i didn't have to do it after all - it was off-the-syllabus stuff
 
  • #15
As Viet told take u = cosx.
Then find du/dxand the relation between both to replace du in place of dx in the integration
 
  • #16
Hmm..use of u=sinx is simpler in this case.
 
  • #17
the only u substitution I've used is in differentiation using the chain rule
 

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