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

The integral of (2sin(2x))/(1+cos(x)) is -4(1+cos(x)-ln|1+cos(x)|)+C

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

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

- 4 (1 + cos (x) - ln ∣ 1 + cos (x) ∣) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot 2 \cdot \int - \frac{u}{1 + u} d u

Take the constant out:

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

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

Apply u-substitution

: \int \frac{v - 1}{v} d v

= 2 \cdot 2 (- \int \frac{v - 1}{v} d v)

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot 2 (- \int 1 - \frac{1}{v} d v)

Apply the Sum Rule:

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

= 2 \cdot 2 (- (\int 1 d v - \int \frac{1}{v} d v))

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 \cdot 2 (- (v - ln ∣ v ∣))

Substitute back

= 2 \cdot 2 (- (1 + cos (x) - ln ∣ 1 + cos (x) ∣))

Simplify

= - 4 (1 + cos (x) - ln ∣ 1 + cos (x) ∣)

Add a constant to the solution

= - 4 (1 + cos (x) - ln ∣ 1 + cos (x) ∣) + C

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What is the integral of (2sin(2x))/(1+cos(x)) ?
The integral of (2sin(2x))/(1+cos(x)) is -4(1+cos(x)-ln|1+cos(x)|)+C