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?


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!

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.