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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

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

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 }}$