What is the integral of (cos(3x))/(sin^2(3x)) ?

The integral of (cos(3x))/(sin^2(3x)) is -1/3 csc(3x)+C

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integral of (cos(3x))/(sin^2(3x))

\int \frac{cos ( 3 x )}{sin ^{2} ( 3 x )} d x

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

Solution steps

\int \frac{cos ( 3 x )}{sin ^{2} ( 3 x )} d x

Use the basic trigonometric identity:

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

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

Apply u-substitution

: \int \frac{cot ( u )}{3 sin ( u )} d u

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x e^{x}, 1 \leq x \leq 7

\frac{\partial}{\partial x} (x^{2} y^{\frac{1}{3}})

\frac{d}{d t} (e^{- t} - e^{- t} t)

f (x) = e^{x c o s (x)}

\frac{x}{x ^{2} - 8}

What is the integral of (cos(3x))/(sin^2(3x)) ?
The integral of (cos(3x))/(sin^2(3x)) is -1/3 csc(3x)+C