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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 2 (2 + cos (x) - 2 ln ∣ 2 + cos (x) ∣) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{v - 2}{v} d v

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

Expand

\frac{v - 2}{v} : 1 - \frac{2}{v}

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

\int 1 d v = v

\int \frac{2}{v} d v = 2 ln ∣ v ∣

= 2 (- (v - 2 ln ∣ v ∣))

Substitute back

= 2 (- (2 + cos (x) - 2 ln ∣ 2 + cos (x) ∣))

Simplify

= - 2 (2 + cos (x) - 2 ln ∣ 2 + cos (x) ∣)

Add a constant to the solution

= - 2 (2 + cos (x) - 2 ln ∣ 2 + cos (x) ∣) + C

\int_{- 5}^{5} \frac{y}{25 - y ^{2}} d y

x^{3} (2 x - 1)^{7}

x \to 0 lim (x + 2)

1 + x

\int x^{3} - 9 x d x

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