<s>[INST] 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!
 [/INST] 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}
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