What is the integral of csc(9x)cot(9x) ?

The integral of csc(9x)cot(9x) is -1/9 csc(9x)+C

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integral of csc(9x)cot(9x)

\int csc (9 x) cot (9 x) d x

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

Solution steps

\int csc (9 x) cot (9 x) d x

Rewrite using trig identities

= \int \frac{cot ( 9 x )}{sin ( 9 x )} d x

Apply u-substitution

: \int \frac{cot ( u )}{9 sin ( u )} d u

= \int \frac{cot ( u )}{9 sin ( u )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{cot ( u )}{sin ( u )} d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{9} (- \frac{1}{v})

Substitute back

= \frac{1}{9} (- \frac{1}{sin ( 9 x )})

Simplify

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

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

Add a constant to the solution

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

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What is the integral of csc(9x)cot(9x) ?
The integral of csc(9x)cot(9x) is -1/9 csc(9x)+C