Question

integral of 1+cos square x /1-cos2x dx

Answer 1

Step-by-step explanation:

We have,

$\int \frac{1 + \cos^{2} (x) }{1 - \cos(2x) }dx$

$= \int \frac{1 + \cos^{2} (x) }{2 \sin^{2} (x) }dx$

$= \frac{1}{2} ( \int \csc^{2} (x) dx + \int \cot^{2} (x) dx)$

As, cot²(θ)=cosec²(θ) - 1 , so,

$= \frac{1}{2}(2 \int \csc^{2} (x) dx - \int \: dx)$

$= - \cot(x) - \frac{1}{2}x + c$