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

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

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

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

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

Solution steps

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

Simplify

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = cos (x)

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

Simplify

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

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