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

The integral of (1+cos(2x))/(sin^2(2x)) is 1/2 (-cot(2x)-csc(2x))+C

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integral of (1+cos(2x))/(sin^2(2x))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1 + cos ( u )}{2 sin ^{2} ( u )} d u

= \int \frac{1 + cos ( u )}{2 sin ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1 + cos ( u )}{sin ^{2} ( u )} d u

Expand

\frac{1 + cos ( u )}{sin ^{2} ( u )} : \frac{1}{sin ^{2} ( u )} + \frac{cos ( u )}{sin ^{2} ( u )}

= \frac{1}{2} \cdot \int \frac{1}{sin ^{2} ( u )} + \frac{cos ( u )}{sin ^{2} ( u )} 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}{2} (\int \frac{1}{sin ^{2} ( u )} d u + \int \frac{cos ( u )}{sin ^{2} ( u )} d u)

\int \frac{1}{sin ^{2} ( u )} d u = - cot (u)

\int \frac{cos ( u )}{sin ^{2} ( u )} d u = - csc (u)

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

Substitute back

u = 2 x

= \frac{1}{2} (- cot (2 x) - csc (2 x))

Add a constant to the solution

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

\frac{\partial}{\partial y} (x + 3 y)

(- 1.3) (3.5)

g (x) = ln^{3} (x)

y = \frac{5 x}{2 ^{x}}, at x = 2

\frac{\partial}{\partial y} (ln (x^{2} + 3 y))

What is the integral of (1+cos(2x))/(sin^2(2x)) ?
The integral of (1+cos(2x))/(sin^2(2x)) is 1/2 (-cot(2x)-csc(2x))+C