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

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

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

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

- tan (x) + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= - tan (x) + x

Add a constant to the solution

= - tan (x) + x + C

(- \frac{10}{63}, \frac{1}{3}), (\frac{2}{3}, - \frac{11}{45})

f (x) = \frac{( 2 - e ^{x} + 2 x )}{3}

\int x y d x

(7, - 7), (5, - 3)

(- 6, - 5), (2, - 10)

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