# Thread: How is it right? (definate integral)

1. ## How is it right? (definate integral)

$\displaystyle \int^{4\pi}_0 \sqrt{1+cos\;x}\;dx$

How is the answer $\displaystyle 8\sqrt{2}$? i keep getting 0!

Thanks

2. Originally Posted by polymerase
$\displaystyle \int^{4\pi}_0 \sqrt{1+cos\;x}\;dx$

How is the answer $\displaystyle 8\sqrt{2}$? i keep getting 0!

Thanks
Take a look at the graph. Since the integrand is never negative the area can't be 0. (The reason is that the square root of a number is never negative.)

-Dan

3. Originally Posted by topsquark
Take a look at the graph. Since the integrand is never negative the area can't be 0. (The reason is that the square root of a number is never negative.)

-Dan
I understand it using a graph but i can't compute it algebraically thou....

4. Originally Posted by polymerase
I understand it using a graph but i can't compute it algebraically thou....
Try
$\displaystyle cos \left ( \frac{1}{2}~x \right ) = \pm \sqrt{\frac{1 + cos(x)}{2}}$

The $\displaystyle \pm$ is chosen by which quadrant your angle is in.

-Dan

5. Originally Posted by polymerase
$\displaystyle \int^{4\pi}_0 \sqrt{1+cos\;x}\;dx$

How is the answer $\displaystyle 8\sqrt{2}$? i keep getting 0!

Thanks
$\displaystyle 1 + \cos x = 2\cos^2 \frac{x}{2}$.

This means, $\displaystyle \int_0^{4\pi} \sqrt{1+\cos x} dx = \sqrt{2} \int_0^{4\pi}\left| \cos \frac{x}{2} \right| ~ dx$

Can you continue?