What is the integral of (cos(4x))/(sin^2(4x)) ?

The integral of (cos(4x))/(sin^2(4x)) is -1/4 csc(4x)+C

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integral of (cos(4x))/(sin^2(4x))

\int \frac{cos ( 4 x )}{sin ^{2} ( 4 x )} d x

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

Solution steps

\int \frac{cos ( 4 x )}{sin ^{2} ( 4 x )} d x

Rewrite using trig identities

= \int cot (4 x) csc (4 x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{cot ( u )}{4 sin ( u )} d u

= \int \frac{cot ( u )}{4 sin ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{cot ( u )}{sin ( u )} d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (cos(4x))/(sin^2(4x)) ?
The integral of (cos(4x))/(sin^2(4x)) is -1/4 csc(4x)+C