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# Calculus Challenge 8!

$\displaystyle R^-_n = \frac{2}{\pi} \int_{0}^{\pi/2}{{(\theta^2 + \ln^2(\cos \theta))}^{-{2}^{-n-1}} \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \cdots + \frac{1}{2}\sqrt{\frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)}}}} \ d\theta} = {(\ln 2)}^{-{2}^{-n}}$ $\displaystyle R^+_n = \frac{2}{\pi} \int_{0}^{\pi/2}{{(\theta^2 + \ln^2(\cos \theta))}^{{2}^{-n-1}} \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \cdots + \frac{1}{2}\sqrt{\frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)}}}} \ d\theta} = {(\ln 2)}^{{2}^{-n}}$

Prove the above 2 AMAZING integrals (I must add that adjective, pardon me!)

Note by Kartik Sharma
1 year, 1 month ago

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@Ishan Singh · 1 year, 1 month ago

Can u post the solution? · 1 year ago

Hint: The infinite surds reminds you of what? $$\cos(x) = \sqrt{\frac{1}{2} + \frac{\cos(2x)}{2}}$$ · 1 year ago

We can simplify the integral to $\displaystyle R^-_n = \frac{2}{\pi} \int_{0}^{\pi/2} {(\theta^2 + \ln^2(\cos \theta))}^{-{2}^{-n-1}} \cdot \cos \left(\dfrac{1}{2^n} \cos^{-1} \left(\sqrt{ \frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)} }\right) \right)$ and

$\displaystyle R^+_n = \frac{2}{\pi} \int_{0}^{\pi/2} {(\theta^2 + \ln^2(\cos \theta))}^{{2}^{-n-1}} \cdot \cos \left(\dfrac{1}{2^n} \cos^{-1} \left(\sqrt{ \frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)} }\right) \right)$

but what to do after that? · 1 year ago