Crazy $\int_0^{\frac\pi{2}}\frac {\cos\left((1-2n)\arcsin\left(\frac{\sin(\theta)}{\sqrt 2}\right)\right)}{\sqrt{1-\frac{\sin^2 \theta}{2}}}d\theta$

Question

$$\int_0^{\frac\pi{2}}\frac {\cos\left((1-2n)\arcsin\left(\frac{\sin(\theta)}{\sqrt 2}\right)\right)}{\sqrt{(1-\frac{\sin^2 \theta}{2})}}d\theta$$

I tried using some sort of substitutions but I think this must have some other way to solve and gave me another different integral and gamma functions and all which now I'm uncertain if it's my cup of tea!

Answer 1

$$\begin{align*}\int_0^{\frac\pi{2}}\frac {\cos\left((1-2n)\arcsin\left(\frac{\sin(\theta)}{\sqrt 2}\right)\right)}{\sqrt{(1-\frac{\sin^2 \theta}{2})}}d\theta & = \frac{\pi}{2}{_2F_1}{(1-n, n;1;\frac 1{2})}\\ & = \frac {\pi}{2}\frac {\sqrt \pi}{\Gamma(1-\frac n{2})\Gamma(\frac 1{2} + \frac n{2})}\\& = \frac {\pi}{2}P_{-n}(1-2x)|_{x = \frac 1{2}}\\ \end{align*}$$

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• In reply to the comment: $\phi = \sin^{-1}(\sqrt x \sin(\theta))$ $$\left|\int_0^{\sin^{-1}\sqrt x} \frac {\cos\left((1-2n)\phi\right)}{\sqrt{(x-\sin^2\phi)}}d\phi = \frac {\pi}{2}{_2F_1}(1-n,n;1;x)\right|_{x = \frac 1{2}}$$

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• #Curiosity $$(1-y^2)^{-1/2}\cos(2n\sin^{-1} y) ={_2F_1}(\frac 1{2}+n, \frac 1{2}-n;\frac 1{2};y^2)$$ $(y, 2n) ≡ (\sqrt x \sin \theta, 1-2n)$ & integrating w.r.t $\theta$ over $(0, \pi/2)$

Answer 2

$$\int_0^{\pi/2}\frac{\cos\left((1-2n)\arcsin\left(\frac{\sin\theta}{\sqrt 2}\right)\right)}{\sqrt{(1-\frac{\sin^2\theta}2)}}\,d\theta$$ Let, $$u=\arcsin\frac{\sin\theta}{\sqrt{2}}\implies du=\frac1{\sqrt{1-\frac{sin^2\theta}2}}\frac{\cos\theta}{\sqrt2}d\theta$$ Multiply and divide by $\cos\theta$ $$\int_0^{\frac\pi2}\frac {\cos\left((1-2n)\arcsin\left(\frac{\sin(\theta)}{\sqrt2}\right)\right)}{\sqrt{(1-\frac{\sin^2\theta}2)}}\frac{\cos\theta}{\cos\theta}\,d\theta$$ And note that if $u=\arcsin\left(\frac{\sin\theta}{\sqrt2}\right)$ then $\cos \theta=\sqrt{1-2\sin^{2} u}$ $$\int_0^{\pi/4}\frac{\cos(u(1-2n))}{\sqrt{1-2\sin^{2} u}}du $$ By the double angle formulas we have $\cos2x=\cos^2x-\sin^2x$. $$\int_0^{\pi/4}\frac{\cos(u(1-2n))}{\sqrt{\cos2u}}\,du$$ I hope you can go from here.

Answer 3

$$I = \int_0^{\frac {\pi}{2}}\frac{\cos{\left((1-2n)\sin^{-1}\left(\sqrt x\sin(\phi)\right)\right)}}{\sqrt{(1-x\sin^2(\phi))}}d\phi$$

$$\begin{align*} (1-x^2)^{-\frac 1{2}}\cos(2n\sin^{-1}x) & = _2F_1(0.5 + n, 0.5-n;0.5;x^2)\\ & = \text{let, }2n =1-2n \text{ & } x = \frac {\sin\phi}{\sqrt 2} \text {integrating both sides w.r.t } \phi \text{ over } \in [0, \pi/2]\\ \int_0^{\frac {\pi}{2}} \frac {\cos{\left((1-2n)(\sin^{-1}\left(\frac {\sin\phi}-{\sqrt 2}\right)\right)}}{\sqrt{(1-\frac {\sin^2\phi}{2})}}d\phi &=\int_0^{\frac {\pi}{2}} {_2F_1(1-n,n;0.5;\frac{\sin^2\phi}{2})}d\phi\\ & =\sum_{k\geq0}\frac {(1-n)_k(n)_k}{(0.5)_kk!2^k}\int_0^{\frac {\pi}{2}}{\sin^{2k}\phi}d\phi\\ & =\sum_{k\geq0}\frac {(1-n)_k(n)_k}{(0.5)_kk!2^k}\beta(0.5,k+0.5)\\ & =\sum_{k\geq0}\frac {(1-n)_k(n)_k}{\color{red}{(0.5)_k}k!2^k}\frac{\sqrt\pi\times \color{red}{\Gamma{(k+0.5)}}}{2\Gamma{(k+1)}}\\ & =\frac {\sqrt\pi}{2}\sum_{k\geq0} {\frac {(1-n)_k(n)_k}{(1)_k}\frac {(0.5)_k}{k!}}\\ & = \frac {\sqrt\pi}{2} {_2F_1}(1-n,n;1;0.5)\\ \end{align*}$$

You can use: $$ {_2F_1}(a, 1-a;b;0.5) = \frac {2^{1-b}\sqrt\pi \Gamma(b)}{\Gamma{(\frac{a+b}{2})}\Gamma{(\frac{b-a+1}{2})}}$$ (I don't know what this Identity is called nor do I remember where I read these all)

Now, Where I found that very first formula and all is other part of the story(I found that by differentiating 0.5sin(2narcsinx)/n = F(0.5+n,0.5-n;0.5;x^2)

If any typo let it be blamed in the name of Dyslexia Cheers:))