# Thread: Integration by Substitution - Cal I problem

1. ## Integration by Substitution - Cal I problem

$\displaystyle \displaystyle\int\frac{\cos\sqrt{x}}{\sqrt{x}}dx$

Let $\displaystyle u=\sqrt{x}=x^{\frac{1}{2}}$

$\displaystyle du =\frac{1}{2\sqrt{x}} dx$

$\displaystyle \displaystyle\int\frac{\cos u}{u} du$

$\displaystyle \displaystyle\int(\cos u)u^{\frac{-1}{2}} du$

I'm new to integrals and to the substitution rule. I don't know how to integrate products yet (besides by multiplying them out first). I'm kinda stuck right here. Can someone please give me a hint to get me unstuck?

Thanks

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

Let $\displaystyle \displaystyle u = \sqrt{x}$ so that $\displaystyle \displaystyle \frac{du}{dx} = \frac{1}{2\sqrt{x}}$.

Then $\displaystyle \displaystyle 2\int{\cos{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\righ t)\,dx} = 2\int{\cos{u}\,\frac{du}{dx}\,dx}$

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

$\displaystyle \displaystyle = 2\sin{u} + C$

$\displaystyle \displaystyle = 2\sin{\sqrt{x}} + C$.

3. Man, I think it is going to take me a while before that comes second nature to me. Thank You

4. The key is to be able to see an "inner" function, and its derivative being multiplied to the "outer" function. If it almost looks right, you can use scalar multiplication to get it to the right form.

5. you might as well take u^2 = x.. if you have some root form to integrate, it is better to substituent^n = x

6. Originally Posted by Prove It
The key is to be able to see an "inner" function, and its derivative being multiplied to the "outer" function. If it almost looks right, you can use scalar multiplication to get it to the right form.
After doing several sets of those types of problems tonight (this morning), its starting to click. Feels good!