<s>[INST] 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 
= $$\int_{1}^{2} \frac{1}{2}u^\frac{1}{2}du$$
= $$\left[\frac{1}{3}u^\frac{3}{2}\right]_1^2$$
= $$\frac{1}{3}\times (2\sqrt3 - 1)$$
= $$\frac{2\sqrt3-1}{3}$$
 [/INST] $$\int \cos(x) \sqrt{2\sin(x)+1} dx = \frac{1}{2}\int \sqrt{2\sin(x)+1}d(2\sin(x)+1) = \frac{1}{2}(\frac{2}{3}(2\sin(x)+1)\sqrt{2\sin(x)+1}+C)=F(x)$$
So the result is $F(\frac{\pi}{6}) - F(0)$
</s>