Problem with definite integral $\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$ $$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
The question says 'evaluate the integral using the suggested substitution. It gives $u=\cos x$. But I think Let $u=1-2\sin x$ is better. 
$$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
$$u=1-2\sin x$$
$$du=-2\cos x dx$$
$$=-\frac{1}{2}\int_{1}^{0}\sqrt{u}du$$
$$=\frac{1}{2}\int_{0}^{1}\sqrt{u}du$$
$$=\left | \frac{u^{\frac{3}{2}}}{3} \right |_{0}^{1}$$
$$=\frac{1}{3}$$
My question is how to solve it by using $u=\cos x$. Can anyone show the solution for it? Thanks a lot!
 A: Well, @Nilan has the better way to go.  But, here is another "Brute Force," double substitution method that works.
Write, $\sin x = \sqrt{1-\cos^2 x}$ (the positive square root is appropriate here since $0\le x\le \frac{\pi}{6}$.  Then, with $u = \cos x$, $du = -\sin x dx=-\sqrt{1-u^2}du$, and the limits of integration extend from $u=1$ to $u=\frac{\sqrt{3}}{2}$.  Thus, 
$$\int_0^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x}dx= \int_{\frac{\sqrt{3}}{2}}^1 \frac{u\sqrt{1-2\sqrt{1-u^2}}}{\sqrt{1-u^2}}du$$
Now, we make a second substitution.  Let $y=\sqrt{1-u^2}$.  Then, $dy=\frac{-2u}{\sqrt{1-u^2}}du$ and the limits of integration go from $y=\frac12$ to $y=0$.  (Note:  This second substitution is identical to making the original substitution $u=\sin x$).  Thus, 
$$\int_0^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x}dx=\int_0^{\frac12} \sqrt{1-2y}dy=\frac13$$
A: Consider the integral
\begin{align}
\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx
\end{align} 
with the substitution $u = \cos(x)$. Making the desired substitution the integral becomes
\begin{align}
I = \int_{\sqrt{3}/2}^{1} u \sqrt{ \frac{1 - 2 \sqrt{1-u^{2}}}{1 - u^{2} } } du
\end{align} 
Now make the substitution $t = 1 - u^{2}$ to obtain the integral
\begin{align}
I = \int_{0}^{1/2} \sqrt{1 - 2t} \, dt.
\end{align}
Making one last change of $y = 1-2t$ which leads to
\begin{align}
I = \frac{1}{2} \int_{0}^{1} \sqrt{y} \, dy = \frac{1}{3}.
\end{align}
A: The substitution is fine, and if you were directed to do so then there is no option, but I think there's a simpler way to go. First, observe that
$$\int\sqrt x\;dx=\frac23 x^{3/2}+C\implies \int f'(x)\sqrt{f(x)}\;dx=\frac23(f(x))^{3/2}+C$$
In our case, we have
$$\cos x=-\frac12(1-2\sin x)'$$
and we thus get directly
$$\int_0^{\pi/6}\cos x\sqrt{1-2\sin x}\;dx=\left.-\frac12\frac23(1-2\sin x)^{3/2}\right|_0^{\pi/6}=$$
$$=-\frac13\left(1-2\sin\frac\pi6-1\right)=\frac23\frac12=\frac13$$
