What is the integral of (sin^3(x))/(2+cos(x)) ?

The integral of (sin^3(x))/(2+cos(x)) is 3ln|cos(x)+2|+1/2 (cos(x)+2)^2-4cos(x)-8+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin^3(x))/(2+cos(x))

\int \frac{sin ^{3} ( x )}{2 + cos ( x )} d x

3 ln ∣ cos (x) + 2 ∣ + \frac{1}{2} (cos (x) + 2)^{2} - 4 cos (x) - 8 + C

Solution steps

\int \frac{sin ^{3} ( x )}{2 + cos ( x )} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{1 - u ^{2}}{2 + u} : \frac{1}{2 + u} - \frac{u ^{2}}{2 + u}

Apply the Sum Rule:

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

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

\int \frac{1}{2 + u} d u = ln ∣ 2 + u ∣

\int \frac{u ^{2}}{2 + u} d u = \frac{( 2 + u ) ^{2}}{2} - 4 (2 + u) + 4 ln ∣ 2 + u ∣

= - (ln ∣ 2 + u ∣ - (\frac{( 2 + u ) ^{2}}{2} - 4 (2 + u) + 4 ln ∣ 2 + u ∣))

Substitute back

u = cos (x)

= - (ln ∣ 2 + cos (x) ∣ - (\frac{( 2 + cos ( x ) ) ^{2}}{2} - 4 (2 + cos (x)) + 4 ln ∣ 2 + cos (x) ∣))

Simplify

- (ln ∣ 2 + cos (x) ∣ - (\frac{( 2 + cos ( x ) ) ^{2}}{2} - 4 (2 + cos (x)) + 4 ln ∣ 2 + cos (x) ∣)) : 3 ln ∣ cos (x) + 2 ∣ + \frac{1}{2} (cos (x) + 2)^{2} - 4 cos (x) - 8

= 3 ln ∣ cos (x) + 2 ∣ + \frac{1}{2} (cos (x) + 2)^{2} - 4 cos (x) - 8

Add a constant to the solution

= 3 ln ∣ cos (x) + 2 ∣ + \frac{1}{2} (cos (x) + 2)^{2} - 4 cos (x) - 8 + C

\int \frac{3 x ^{2}}{1 + x ^{2}} d x

\int \frac{5}{4 x ^{2} - 4 x + 10} d x

cos^{4} (x)

\int \frac{e ^{x} + 4}{e ^{x}} d x

\int \frac{3 x + 1}{x} d x

What is the integral of (sin^3(x))/(2+cos(x)) ?
The integral of (sin^3(x))/(2+cos(x)) is 3ln|cos(x)+2|+1/2 (cos(x)+2)^2-4cos(x)-8+C