Integration

$\large \displaystyle I(n) = \int_0^{\pi} \ln(1 - 2n \cos(x) + n^2) dx$

Since $\color{#20A900} \int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ :

$\large \displaystyle I(n) = \frac12 \left [ \int_0^{\pi} \ln(1 - 2n \cos(x) + n^2) dx + \int_0^{\pi} \ln(1 - 2n \cos(0 + \pi - x) + n^2) dx \right ]$

$\large \displaystyle I(n) = \frac12 \left [ \int_0^{\pi} \ln(1 - 2n \cos(x) + n^2) dx + \int_0^{\pi} \ln(1 + 2n \cos(x) + n^2) dx \right ]$

$\large \displaystyle I(n) = \frac12 \left [ \int_0^{\pi} \ln (1 + 2n^2 + n^4 - 4n^2 \cos^2(x) ) dx \right ]$

$\large \displaystyle I(n) = \frac12 \left [ \int_0^{\pi} \ln (1 + 2n^2 (1- 2 \cos^2(x)) + n^4 ) dx \right ]$

$\large \displaystyle I(n) = \frac12 \left [ \int_0^{\pi} \ln (1 - 2n^2 \cos(2x) + n^4 ) dx \right ]$

$\color{#20A900} \large \displaystyle u = 2x \rightarrow dx = \frac12 du$

$\large \displaystyle I(n) = \frac14 \left [ \int_0^{2 \pi} \ln (1 - 2n^2 \cos(u) + n^4 ) du \right ]$

$\large \displaystyle I(n) = \frac14 \left [ \int_0^{\pi} \ln (1 - 2n^2 \cos(u) + n^4 ) du + \int_{\pi}^{2 \pi} \ln (1 - 2n^2 \cos(u) + n^4 ) du \right ]$

On second integral:

$\color{#20A900} \large \displaystyle v = 2 \pi - u \rightarrow du = - dv$

$\large \displaystyle I(n) = \frac14 \left [ \int_0^{\pi} \ln (1 - 2n^2 \cos(u) + n^4 ) du - \int_{\pi}^{0} \ln (1 - 2n^2 \cos(v) + n^4 ) dv \right ]$

$\large \displaystyle I(n) = \frac14 \left [ \int_0^{\pi} \ln (1 - 2n^2 \cos(u) + n^4 ) du + \int_{0}^{\pi} \ln (1 - 2n^2 \cos(v) + n^4 ) dv \right ]$

$\large \displaystyle I(n) = \frac14 \left [ I(n^2) + I(n^2) \right ]$

$\large \displaystyle I(n) = \frac12 I(n^2)$

$\color{#20A900} \boxed { \large \displaystyle \frac{I(n^2)}{I(n)} = 2 }$

So:

$\color{#3D99F6} \large \displaystyle \frac{I(100)}{I(10)} \cdot \frac{I(36)}{I(6)} = \frac{I(10^2)}{I(10)} \cdot \frac{I(6^2)}{I(6)} = 2\cdot 2 = \boxed {\large \displaystyle 4}$

We can evaluate the integrals, rather than just find relationships between them. Note that for $\mu > 0$ , $\begin{aligned} F(\mu) & = \int_0^\pi \sin^{\mu-1}x\,dx \; = \; 2^{\mu-1}\int_0^\pi \sin^{\mu-1}\tfrac12x \cos^{\mu-1}\tfrac12x\,dx \; = \; 2^\mu \int_0^{\frac12\pi} \sin^{\mu-1}x \cos^{\mu-1}x\,dx \\ & = 2^{\mu-1}B\big(\tfrac12\mu,\tfrac12\mu\big) \; = \; \frac{2^{\mu-1}\Gamma(\tfrac12\mu)^2}{\Gamma(\mu)} \end{aligned}$ so that $F'(\mu) \; = \; F(\mu)\big[\ln2 + \psi\big(\tfrac12\mu\big) - \psi(\mu)\big]$ and hence $\int_0^\pi \ln(\sin x)\,dx \; = \; F'(1) \; = \; F(1)\big[\ln2 + \psi(\tfrac12) - \psi(1)\big] \; = \; \pi\big[\ln2 - \gamma - \ln4 + \gamma\big] \; = \; -\pi\ln2$ and hence we obtain the moderately standard integral $\int_0^\pi \ln(2\sin x)\,dx \; = \; 0$ If we now define $I(\alpha) \; = \; \int_0^\pi \ln(1 - 2\alpha \cos x + \alpha^2)\,dx \; = \; \tfrac12\int_0^{2\pi} \ln(1 - 2\alpha \cos x + \alpha^2)\,dx \hspace{1cm} \alpha \ge 1$ then $I(1) \; = \; \tfrac12\int_0^{2\pi}\ln\big(4\sin^2\tfrac12x\big)\,dx \; = \; \int_0^{2\pi} \ln\big(2\sin\tfrac12x\big)\,dx \; = \; 2\int_0^\pi \ln(2\sin x)\,dx \; = \; 0$ Moreover the substitution $z = e^{ix}$ and some contour integration yields $\begin{aligned} I'(\alpha) & = \tfrac12\int_0^{2\pi} \frac{2\alpha - 2\cos x}{1 - 2\alpha \cos x + \alpha^2} \; = \; \tfrac12 \int_{|z|=1} \frac{2\alpha - z - z^{-1}}{1 - \alpha(z + z^{-1}) + \alpha^2}\,\frac{dz}{i z} \\ & = -\frac{1}{2i}\int_{|z|=1} \frac{2\alpha z - z^2 - 1}{z(z-\alpha)(\alpha z - 1)}\,dz \; = \; -\pi\Big(\mathrm{Res}_{z=0} + \mathrm{Res}_{z=\alpha^{-1}}\Big) \frac{2\alpha z - z^2 - 1}{z(z-\alpha)(\alpha z - 1)} \\ & = -\pi\Big(-\frac{1}{\alpha} + \frac{1 - \alpha^{-2}}{\alpha^{-1} - \alpha}\Big) \; = \; \frac{2\pi}{\alpha} \end{aligned}$ for $\alpha > 1$ , so that $I(\alpha) \; = \; I(1) + 2\pi\ln\alpha \; = \; 2\pi\ln\alpha \hspace{2cm} \alpha \ge 1$ so that $\frac{I(\alpha^2)}{I(\alpha)} \; =\; 2 \hspace{2cm} \alpha \ge 1$ making the answer to the question equal to $\boxed{4}$ .

I'm a bit busy these days . this problem is from AIITS 2 Of FIITJEE.

Try proving I(n^2)/I(n) = 2 by using a+b-x property and adding

You may also approach by writting (1-2ncos(x) + $n^{2}$ ) = (n- $e^{x}$ )(n- $e^{-x}$ )