$$ \int \cos x \sqrt{\cos2x}dx $$

I know that $\cos 2x = 1-2\sin^2x,$ but not sure if it will help me?

Integrate by parts?

$$ \sin x \sqrt{1-2\sin^{2}x}-\int \sin x \frac{-\sin2x}{\sqrt{\cos2x}}dx $$ This doesn't seem to get me anywhere?


Hint: write cos2x = 1 - 2(sinx)^2, and let u = sinx, then du = cosxdx. You can take it from here.


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