What is the integral of (cos^3(3x))/(sin^5(3x)) ?

The integral of (cos^3(3x))/(sin^5(3x)) is -1/12 cot^4(3x)+C

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integral of (cos^3(3x))/(sin^5(3x))

\int \frac{cos ^{3} ( 3 x )}{sin ^{5} ( 3 x )} d x

- \frac{1}{12} cot^{4} (3 x) + C

Solution steps

\int \frac{cos ^{3} ( 3 x )}{sin ^{5} ( 3 x )} d x

Apply u-substitution

: \int \frac{cos ^{3} ( u )}{3 sin ^{5} ( u )} d u

= \int \frac{cos ^{3} ( u )}{3 sin ^{5} ( u )} 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{cos ^{3} ( u )}{sin ^{5} ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{cot ^{3} ( u )}{sin ^{2} ( u )} d u

Use the basic trigonometric identity:

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

= \frac{1}{3} \cdot \int cot^{3} (u) csc^{2} (u) d u

Apply u-substitution

: \int - v^{3} d v

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

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{3} (- \frac{v ^{4}}{4})

Substitute back

= \frac{1}{3} (- \frac{cot ^{4} ( 3 x )}{4})

Simplify

\frac{1}{3} (- \frac{cot ^{4} ( 3 x )}{4}) : - \frac{1}{12} cot^{4} (3 x)

= - \frac{1}{12} cot^{4} (3 x)

Add a constant to the solution

= - \frac{1}{12} cot^{4} (3 x) + C

\int - \frac{1}{2 - x} d x

f (x) = cot^{5} (x)

(x^{2} - y^{2}) d x + x y d y = 0

\int_{7}^{13} \frac{y}{y ^{2} - 3 y - 4} d y

f^{^{'''}} (x) = 6 (x^{- 2})

What is the integral of (cos^3(3x))/(sin^5(3x)) ?
The integral of (cos^3(3x))/(sin^5(3x)) is -1/12 cot^4(3x)+C