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How can we integrate:

$$\int \frac{\sec(x) \, dx}{\sqrt{\cos(2x+\alpha)+\cos(\alpha)}}$$

I used the identity $\cos(C)+\cos(D)$ to obtain $$\int \frac{\sec(x) \, dx}{\sqrt{2\cos(x+\alpha) \, \cos( x)}}$$

But it doesn't seem to help. How to proceed?

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You're on the right track, now multiply by $\sec x$ in the numerator and denominator to get $$\int \frac{(\sec^2 x)dx}{\sqrt \frac{2\cos(x+\alpha)}{\cos x}}$$ Now use: $$\cos(x+\alpha)=\cos x\cos \alpha - \sin x\sin \alpha$$ and you get $$\int \frac{(\sec^2 x)dx}{\sqrt {2(\cos\alpha-\tan x \sin \alpha)}}$$ Now substitute $2(\cos\alpha-\tan x \sin \alpha) = t$ and you're done!

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