What is the integral of (sin(3x))/([1+cos(3x)]^4) ?

The integral of (sin(3x))/([1+cos(3x)]^4) is 1/(9(1+cos(3x))^3)+C

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integral of (sin(3x))/([1+cos(3x)]^4)

\int \frac{sin ( 3 x )}{[ 1 + cos ( 3 x ) ] ^{4}} d x

\frac{1}{9 ( 1 + cos ( 3 x ) ) ^{3}} + C

Solution steps

\int \frac{sin ( 3 x )}{[ 1 + cos ( 3 x ) ] ^{4}} d x

Apply u-substitution

: \int - \frac{1}{3 u ^{4}} d u

= \int - \frac{1}{3 u ^{4}} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= - \frac{1}{3} (- \frac{1}{3 u ^{3}})

Substitute back

u = 1 + cos (3 x)

= - \frac{1}{3} (- \frac{1}{3 ( 1 + cos ( 3 x ) ) ^{3}})

Simplify

- \frac{1}{3} (- \frac{1}{3 ( 1 + cos ( 3 x ) ) ^{3}}) : \frac{1}{9 ( 1 + cos ( 3 x ) ) ^{3}}

= \frac{1}{9 ( 1 + cos ( 3 x ) ) ^{3}}

Add a constant to the solution

= \frac{1}{9 ( 1 + cos ( 3 x ) ) ^{3}} + C

3 x^{2} - 5 x + 1

x \to 0 lim (3 sin (x) - 2)

\int sin (3 x) \cdot cos (3 x) d x

\int e^{y} - 1 d y

\frac{d}{d x} ((x - \frac{1}{x})^{\frac{5}{2}})

What is the integral of (sin(3x))/([1+cos(3x)]^4) ?
The integral of (sin(3x))/([1+cos(3x)]^4) is 1/(9(1+cos(3x))^3)+C