I need help again!

• Jul 30th 2006, 04:50 AM
margaritas
I need help again!
Okay, this question is making me confused.

(i) State the derivative of sin(x^2).

I got that correct but I can't get the next part.

(ii) Find (integrate) x^3 cos (x^2) dx.

Answer: 1/2 [x^2 sin (x^2) + cos (x^2)] + C.

Thanks in advance if you could help!
• Jul 30th 2006, 05:57 AM
galactus
This one can be a little tricky. Use integration by parts.

Let $u=\frac{x^{2}}{2},\;\ dv=2xcos(x^{2})dx,\;\ du=xdx,\;\ v=sin(x^{2})$

Therefore:

$\int{udv}=uv-\int{vdu}$

$\int\frac{x^{2}}{2}(2xcos(x^{2}))dx=\frac{x^{2}}{2 }sin(x^{2})-\int{xsin(x^{2})}dx$

Let $u=x^{2},\;\ \frac{du}{2}=xdx$

The integral portion becomes $\frac{1}{2}\int{sin(u)}du$

Can you finish up now?.
• Jul 30th 2006, 06:49 AM
margaritas
Yay thanks so much, galactus, I finally got the answer!
• Jul 30th 2006, 08:36 AM
topsquark
Quote:

Originally Posted by galactus
This one can be a little tricky. Use integration by parts.

Let $u=\frac{x^{2}}{2},\;\ dv=2xcos(x^{2})dx,\;\ du=xdx,\;\ v=sin(x^{2})$

Therefore:

$\int{udv}=uv-\int{vdu}$

$\int\frac{x^{2}}{2}(2xcos(x^{2}))dx=\frac{x^{2}}{2 }sin(x^{2})-\int{xsin(x^{2})}dx$

Let $u=x^{2},\;\ \frac{du}{2}=xdx$

The integral portion becomes $\frac{1}{2}\int{sin(u)}du$

Can you finish up now?.

Actually, I find it a bit easier to substitute first:

Let $y = x^2$

Then:
$\int dx \, x^3 cos(x^2) = \frac{1}{2} \int (2xdx) x^2 cos(x^2) = \frac{1}{2} \int dy \, y cos(y)$

Then you only need integrate by parts once. (I hate integration by parts so I only do as many as I have to.)

-Dan
• Jul 31st 2006, 05:20 AM
margaritas
Thanks for the alternative, Dan. I will try your method when I come across a similar question. :)