What is the integral of (1-cos(2x))/(1+cos(2x)) ?

The integral of (1-cos(2x))/(1+cos(2x)) is -x+tan(x)+C

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integral of (1-cos(2x))/(1+cos(2x))

\int \frac{1 - cos ( 2 x )}{1 + cos ( 2 x )} d x

- x + tan (x) + C

Solution steps

\int \frac{1 - cos ( 2 x )}{1 + cos ( 2 x )} d x

\frac{1 - cos ( 2 x )}{1 + cos ( 2 x )} = tan^{2} (x)

= \int tan^{2} (x) d x

Rewrite using trig identities

= \int - 1 + sec^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d x + \int sec^{2} (x) d x

\int 1 d x = x

\int sec^{2} (x) d x = tan (x)

= - x + tan (x)

Add a constant to the solution

= - x + tan (x) + C

\int_{0}^{5} 5 - y d y

\frac{t}{t ^{2} + 7}

\frac{d x}{d t} + 4 t x = t, x (0) = 0

\int \frac{x}{7} 49 + x^{2} d x

\frac{d}{d x} (e^{- 2 x} - x)

What is the integral of (1-cos(2x))/(1+cos(2x)) ?
The integral of (1-cos(2x))/(1+cos(2x)) is -x+tan(x)+C