# Thread: cos(SQRT{x}) dx (some work included).

1. ## cos(SQRT{x}) dx (some work included).

Hello guys here is my problem uploaded: , see the attached image of my scanned paper

The thing that i wonder about here is the hint : 2\$cos u * u * dx-

Where this this expression come from, how did they apply Cos to .where did it suddenly come from. I can understand why the constant was suddenly omitted from the right side of the equation (Constant Multiplie Rule). Would be happy if someone could solve this for me.

2. Let me just retype what you have so others can read it easier:

$\displaystyle \int \cos \sqrt x\, dx$

Substitue $x = u^2$
$dx = 2u\,du$

$\displaystyle \int \cos \sqrt x\, dx = \int \cos u\, 2u\, du = 2\int u\cos u\, du$

Is this where you are stuck?

Try to do integration by parts.

3. We won't solve it for you, but the trick is to try to get the derivative of $\displaystyle \sqrt{x}$, i.e. $\displaystyle \frac{1}{2\sqrt{x}}$, as a factor.

So $\displaystyle \int{\cos{\sqrt{x}}\,dx} = \int{\frac{2\sqrt{x}\cos{\sqrt{x}}}{2\sqrt{x}}\,dx }$

$\displaystyle = 2\int{\sqrt{x}\cos{(\sqrt{x})}\,\frac{1}{2\sqrt{x} }\,dx}$.

Now make the substitution $\displaystyle u = \sqrt{x}$ so that $\displaystyle \frac{du}{dx}= \frac{1}{2\sqrt{x}}$ and the integral becomes

$\displaystyle 2\int{u\cos{u}\,\frac{du}{dx}\,dx}$

$\displaystyle = 2\int{u\cos{u}\,du}$.

You can now solve this using integration by parts.

4. Snowteas and prove its ways of solving are different, Snowtea you based your on my variable substitution by using u^2.

Prove it could you elaborate on your expression of getting to: \displaystyle \int{\cos{\sqrt{x}}\,dx} = \int{\frac{2\sqrt{x}\cos{\sqrt{x}}}{2\sqrt{x}}\,dx } because all I can understand now after looking at the derivate of sqrt[x] is that the derivate is given by using "Difference of two squares" formula and then you get 1/2*sqrt[x]. Thats understandable after trying it out myself half an hour ago.

Also both of your answers lead to the same integral, here is where i am stuck again.

I have integrated by using the DI shorcut (http://www.delmar.edu/math/MLC/Forms...By%20Parts.pdf) for fast results now.

Here is my solution Tell me if it's appropriate or not.

5. Riazy, your solution is really hard to read.
Can you try to retype it in latex?

For example typing
$$\displaystyle \int \frac{x^2}{\cos(\sqrt{x})} dx$$
gives
$\displaystyle \int \frac{x^2}{\cos(\sqrt{x})} dx$.

6. What Prove It did, was simply multiply by 1.

$\displaystyle \int{\cos{\sqrt{x}}\,dx} = \displaystyle \int{\cos{\sqrt{x}} \times 1 \,dx} = \displaystyle \int{\cos{\sqrt{x}} \times \dfrac{2\sqrt{x}}{2\sqrt{x}}\,dx} = \int{\frac{2\sqrt{x}\cos{\sqrt{x}}}{2\sqrt{x}}\,dx }$

EDIT:

And no, that's not it. Integration by parts:

$\displaystyle \int uv' dx = [uv] - \int vu' \ dx$

Hence;

$\displaystyle \int u\cos u\ du = [u \sin u] - \int \sin u \cdot 1\ du$

Can you complete it now?