Compute the following integral (u-sub):

Question

anonymous · Answer

$$\int\limits_{}^{}\frac{ x \cos (x^2)  }{ \sqrt{\sin (x^2)}}dx$$

hartnn · Answer

did you try u=x^2 first?

anonymous · Answer

I did. I ended up with x + C, which was marked as incorrect.
Maybe it has something to do with tangent? I am rather confused by this question.

hartnn · Answer

x+c ? how? 
first put x^2 =u 
next put sin u = t 
or you can directly first put 
sin x^2 = u 
du=... ?

anonymous · Answer

I attempted to put u = x^2 and v = sin(u)... perhaps that was my mistake.
If sin x^2 = u, du/2 = xcos x^2dx?

hartnn · Answer

yes!
that would totally simplify it...

anonymous · Answer

Well, here is what I did... Maybe my mistake was in computation.

hartnn · Answer

$\int du/(2\sqrt u)$

anonymous · Answer

1$$\frac{ 1 }{ 2 } \int\limits_{}^{} v^{-1/2}dv = \frac{ v^{-1/2+1} }{ 1/2 } + C = \sqrt{v}+C=\sqrt{\sin(u)}+C=\sqrt{x^2}+C=x+C$$

anonymous · Answer

Where u=x^2 and v=sin(u)

hartnn · Answer

$\sqrt{\sin x^2}+c$
is correct final answer.

anonymous · Answer

Ah, I see.$$\sqrt{\sin(u)}+C \neq \sqrt{x^2}+C.$$ So would the answer be $$\sqrt{\sin(x^2)}+C?$$

anonymous · Answer

I see... Can that be simplified any further?

hartnn · Answer

i don't think so....no

anonymous · Answer

Okay. I see what I did wrong now... thank you for your help again, though I'll  probably need it again soon! 
Much appreciated!

hartnn · Answer

welcome ^_^