U-substitution integration help

[bubbles]

Homework Statement

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

 

[radou]

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

 

[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

 

[courtrigrad]

Let u = \frac{1}{2}x. Then du = \frac{1}{2} dx.So you have: 2 \int \cos u \; du

 

[radou]

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

Yeah I'm confused. The integrand is equal to

2/sqrt(2)*x so the (1/sqrt(2))x^2 + C

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.

 

[bubbles]

I think I got the right answer:

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

\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

But I don't understand how you got that.

 

[radou]

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 .

 

[bubbles]

Thank you , radou. I understand it now.

Nevermind what I said in post #6; I just realize that nealh149 was talking about another problem.