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### $\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + 1 + \sin x}}{{1 + \cos 2x}}} dx = ?$

The given integral can be split as,

$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}} dx + \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{1 + \cos 2x}}} dx$

$\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}$ is an odd function. Thus, the given integral

$= 0 + \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{2{{\cos }^2}x}}} dx = \frac{1}{2}\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{{\sec }^2}xdx} = \frac{1}{2}\left. {\tan x} \right|_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} = \frac{1}{{\sqrt 3 }}$

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$