What is the integral of (cos(19x))/(sin^2(19x)) ?

The integral of (cos(19x))/(sin^2(19x)) is -1/19 csc(19x)+C

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integral of (cos(19x))/(sin^2(19x))

\int \frac{cos ( 19 x )}{sin ^{2} ( 19 x )} d x

- \frac{1}{19} csc (19 x) + C

Solution steps

\int \frac{cos ( 19 x )}{sin ^{2} ( 19 x )} d x

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

= \int \frac{cot ( 19 x )}{sin ( 19 x )} d x

Apply u-substitution

: \int \frac{cot ( u )}{19 sin ( u )} d u

= \int \frac{cot ( u )}{19 sin ( u )} d u

Take the constant out:

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

= \frac{1}{19} \cdot \int \frac{cot ( u )}{sin ( u )} d u

Rewrite using trig identities

= \frac{1}{19} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{19} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{19} (- \frac{1}{v})

Substitute back

= \frac{1}{19} (- \frac{1}{sin ( 19 x )})

Simplify

\frac{1}{19} (- \frac{1}{sin ( 19 x )}) : - \frac{1}{19} csc (19 x)

= - \frac{1}{19} csc (19 x)

Add a constant to the solution

= - \frac{1}{19} csc (19 x) + C

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What is the integral of (cos(19x))/(sin^2(19x)) ?
The integral of (cos(19x))/(sin^2(19x)) is -1/19 csc(19x)+C