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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

= \int csc^{4} (3 x) cot (3 x) d x

Apply u-substitution

: \int csc^{4} (u) cot (u) \frac{1}{3} d u

= \int csc^{4} (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^{4} (u) cot (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{cos ( u )}{sin ^{5} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{5}} d v

= \frac{1}{3} \cdot \int \frac{1}{v ^{5}} d v

Apply the Power Rule

: - \frac{1}{4 v ^{4}}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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