Calculating $\int_{0}^{\frac{\pi}{6}} \cos{(x)} \sqrt{2\sin (x)+1} dx$ I am trying to calculate the value of the following
 $$\int_{0}^{\frac{\pi}{6}} cosx  \sqrt{2sin x+1} dx$$
I used a substitution method. 
$$u = 2 \sin (x) + 1$$
$$\frac{du}{dx} = 2\cos (x)$$
$$\frac{u}{2 \cos (x)}du = dx$$
hence 
$$\int_{0}^{\frac{\pi}{6}} \cos (x)  \sqrt{2 \sin (x)+1} dx$$ = 
$$\int_{1}^{2} \cos (x)  \sqrt{u} \times\frac{u}{2 \cos (x)}du$$
= $$\int_{1}^{2} \frac{1}{2}u^\frac{3}{2}du$$
= $$\left[\frac{1}{5}u^\frac{5}{2}\right]_1^2$$
but I can't seem to move any further. 
Can anyone help please?
Many thanks. 
UPDATE
The third line is incorrect. It should be
$$\frac{du}{2 \cos (x)} = dx$$
hence 
= $$\int_{1}^{2} \frac{1}{2}u^\frac{1}{2}du$$
= $$\left[\frac{1}{3}u^\frac{3}{2}\right]_1^2$$
= $$\frac{1}{3}\times (2\sqrt3 - 1)$$
= $$\frac{2\sqrt3-1}{3}$$

You had this:
$$\frac{du}{dx}=2\cos(x)$$
You should've moved the $dx$ to the other side to get
$$du=2\cos(x)\ dx$$
but you had an extra $u$.  Follow this, and the rest of your work is fine.
