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

[karush]

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

 

[MarkFL]

Try the substitution:

$$u=\cos(x)$$

Now you should get something you can integrate directly. :D

 

[Jameson]

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

 

[karush]

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

 

[MarkFL]

so then

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

kinda easy... if correct

You have a sign error...

And then...don't forget the constant of integration and then back-substitute for $u$. :D

 

[karush]

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

 

[MarkFL]

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

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