What is the integral of (cos(3x))/(sin^4(3x)) ?

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

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integral of (cos(3x))/(sin^4(3x))

\int \frac{cos ( 3 x )}{sin ^{4} ( 3 x )} d x

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

Solution steps

\int \frac{cos ( 3 x )}{sin ^{4} ( 3 x )} d x

Rewrite using trig identities

= \int \frac{cot ( 3 x )}{sin ^{3} ( 3 x )} d x

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= \int csc^{3} (3 x) cot (3 x) d x

Apply u-substitution

: \int csc^{3} (u) cot (u) \frac{1}{3} d u

= \int csc^{3} (u) cot (u) \frac{1}{3} 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 csc^{3} (u) cot (u) d u

Apply u-substitution

: \int - v^{2} d v

= \frac{1}{3} \cdot \int - v^{2} d v

Take the constant out:

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

= \frac{1}{3} (- \int v^{2} d v)

Apply the Power Rule

: \frac{v ^{3}}{3}

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

Substitute back

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

Simplify

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

= - \frac{1}{9} csc^{3} (3 x)

Add a constant to the solution

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

f (x) = 2 x^{2} - 18 x + 5, at x = 0

x \to \infty lim (\frac{3 x ^{2} - x - 2}{5 x ^{2} - 4 x + 1})

f (x) = \frac{x ^{2}}{3 + 8 x}

(\frac{x ^{2}}{2})^{^{'}}

\int_{0}^{4} x 17 - x^{2} d x

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