What is the integral of (1-cos(4y))/(1+cos(4y)) ?

The integral of (1-cos(4y))/(1+cos(4y)) is 1/2 (-cot(4y)+csc(4y)-2y)+C

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integral of (1-cos(4y))/(1+cos(4y))

\int \frac{1 - cos ( 4 y )}{1 + cos ( 4 y )} d y

\frac{1}{2} (- cot (4 y) + csc (4 y) - 2 y) + C

Solution steps

\int \frac{1 - cos ( 4 y )}{1 + cos ( 4 y )} d y

Apply u-substitution

= \int \frac{1 - cos ( u )}{4 ( 1 + cos ( u ) )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1 - cos ( u )}{1 + cos ( u )} d u

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

= \frac{1}{4} \cdot \int - \frac{2 cos ( u )}{1 + cos ( u )} + 1 d u

Apply the Sum Rule:

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

= \frac{1}{4} (- \int \frac{2 cos ( u )}{1 + cos ( u )} d u + \int 1 d u)

\int \frac{2 cos ( u )}{1 + cos ( u )} d u = 2 (- csc (u) + cot (u) + u)

\int 1 d u = u

= \frac{1}{4} (- 2 (- csc (u) + cot (u) + u) + u)

Substitute back

u = 4 y

= \frac{1}{4} (- 2 (- csc (4 y) + cot (4 y) + 4 y) + 4 y)

Simplify

\frac{1}{4} (- 2 (- csc (4 y) + cot (4 y) + 4 y) + 4 y) : \frac{1}{2} (- cot (4 y) + csc (4 y) - 2 y)

= \frac{1}{2} (- cot (4 y) + csc (4 y) - 2 y)

Add a constant to the solution

= \frac{1}{2} (- cot (4 y) + csc (4 y) - 2 y) + C

What is the integral of (1-cos(4y))/(1+cos(4y)) ?
The integral of (1-cos(4y))/(1+cos(4y)) is 1/2 (-cot(4y)+csc(4y)-2y)+C