# U-substitution integration help

1. Nov 29, 2006

### bubbles

1. The problem statement, all variables and given/known data

$$\int cos(1/2)x = 2sin(x/2) + c$$ ==> Correct Answer

This is an answer to one of my homework problems, but I don't understand why the indefinite integtral of cos(1/2)x is that. I've learned u-substitution and the power rule, but using those methods, I got:

2. The attempt at a solution

$$u = .5x, du = .5dx \int cos(u)du = sin(u)+c = sin(.5x)+c$$ (wrong answer?)

2. Nov 29, 2006

You made a mistake while substituting. $$du = \frac{1}{2}dx$$ implies $$dx = 2 du$$.

3. Nov 29, 2006

### nealh149

$$\int cos(1/2)x = 2sin(x/2) + c$$

Yeah I'm confused. The integrand is equal to

2/sqrt(2)*x so the integral is equal to (1/sqrt(2))x^2 + C

Last edited: Nov 29, 2006
4. Nov 29, 2006

Let $$u = \frac{1}{2}x$$. Then $$du = \frac{1}{2} dx$$.

So you have: $$2 \int \cos u \; du$$

Last edited: Nov 29, 2006
5. Nov 29, 2006

What is there to be confused about? And where did you get these 'sqrt-s' from? As said before, you only have to substitute dx = 2 du and u = 1/2 x into the original integral, in order to obtain $$\int 2 \cos u du$$, which I assume you'll know how to handle.

6. Nov 29, 2006

### bubbles

I think I got the right answer:

$$\int cos(1/2)x dx = 2 \int cos(.5x)(.5) + c = 2sin(.5x) + c$$

But I don't understand how you got that.

7. Nov 29, 2006

OK, again: $$\int \cos(\frac{1}{2}x)$$ = after substituting = $$\int 2 \cos u du = 2 \sin u + C = 2 \sin (\frac{1}{2}x) + C$$ .