What is the integral of (cos(4x))/(sin^3(4x)) ?

The integral of (cos(4x))/(sin^3(4x)) is -1/8 cot^2(4x)+C

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integral of (cos(4x))/(sin^3(4x))

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

- \frac{1}{8} cot^{2} (4 x) + C

Solution steps

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

= \int csc^{2} (4 x) cot (4 x) d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} 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 u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = cot (4 x)

= - \frac{1}{4} \cdot \frac{cot ^{2} ( 4 x )}{2}

Simplify

- \frac{1}{4} \cdot \frac{cot ^{2} ( 4 x )}{2} : - \frac{1}{8} cot^{2} (4 x)

= - \frac{1}{8} cot^{2} (4 x)

Add a constant to the solution

= - \frac{1}{8} cot^{2} (4 x) + C

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What is the integral of (cos(4x))/(sin^3(4x)) ?
The integral of (cos(4x))/(sin^3(4x)) is -1/8 cot^2(4x)+C