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

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

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

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

- \frac{1}{2} ln (2 ∣ cos (x) ∣^{2}) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + cos (2 x)

= - \frac{1}{2} ln ∣ 1 + cos (2 x) ∣

Simplify

- \frac{1}{2} ln ∣ 1 + cos (2 x) ∣ : - \frac{1}{2} ln (2 ∣ cos (x) ∣^{2})

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

Add a constant to the solution

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

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\int_{1}^{2} e^{\frac{1}{x}} d x

\int (\frac{2 y}{( x ^{2} + y ^{2} ) ^{2}}) d y

f (x) = 4 x^{- \frac{3}{8}}

\frac{d}{d x} (\frac{x ^{7}}{7} - x^{3} - \frac{3}{x} + \frac{1}{5 x ^{5}})

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