Finding the Anti-Derivative of sin(x)cos(cos(x))

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    Integral
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Discussion Overview

The discussion revolves around evaluating the integral $\displaystyle \int{\sin{x}\cos{\left(\cos{x}\right)}}dx$. Participants explore various substitution methods and share their thoughts on the integration process, focusing on the anti-derivative of the given function.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to approach the integral, suggesting a possible relationship between $\cos{\left(\cos{x}\right)}$ and the derivative of $\sin{x}$.
  • Several participants propose the substitution $u=\cos(x)$, with one noting that this leads to a more straightforward integral.
  • Another participant reiterates the substitution and provides the differential $du=-\sin(x)dx$ as part of their response.
  • There is a correction regarding a sign error in the integration result, with a participant emphasizing the importance of including the constant of integration and back-substituting for $u$.
  • Participants confirm the integration result as $-\sin(u) + C$, with a focus on the need to revert to the original variable $x$.

Areas of Agreement / Disagreement

Participants generally agree on the substitution method and the resulting integral, but there is a noted correction regarding a sign error. The discussion remains focused on the integration process without reaching a final consensus on the complete solution.

Contextual Notes

Some participants mention the need to back-substitute for $u$, indicating that the discussion is still in the process of finding the anti-derivative in terms of the original variable.

karush
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Evaluate the Integral
$
\displaystyle
\int{\sin{x}\cos{\left(\cos{x}\right)}}dx
$
I am clueless with this; I thot maybe the $\cos{\left(\cos{x}\right)}$(Tauri)
would be somehow be a $d/dx$ of $\sin{x}$ but perhaps there is another way
 
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Try the substitution:

$$u=\cos(x)$$

Now you should get something you can integrate directly. :D
 
Try this substitution: $u=\cos(x)$, $du=-\sin(x)dx$.

EDIT: Ok, Mark beat me but I want to keep this post since I started replying before he posted! :p
 
Jameson said:
Try this substitution: $u=\cos(x)$, $du=-\sin(x)dx$.

EDIT: Ok, Mark beat me but I want to keep this post since I started replying before he posted! :p

so then

$\displaystyle
-\int \cos{\left(u\right)} du = \sin{\left(u\right)}
$

kinda easy... if correct:cool:
 
karush said:
so then

$\displaystyle
-\int \cos{\left(u\right)} du = \sin{\left(u\right)}
$

kinda easy... if correct:cool:

You have a sign error...

And then...don't forget the constant of integration and then back-substitute for $u$. :D
 
$\displaystyle -\int \cos{\left(u\right)} du = -\sin{\left(u\right)}+C$
 
karush said:
$\displaystyle -\int \cos{\left(u\right)} du = -\sin{\left(u\right)}+C$

Now you want to back-substitute for $u$ so that you have the anti-derivative in terms of the original variable $x$.
 

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