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

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

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

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

\frac{1}{4} (- \frac{1}{3} csc^{3} (4 x) + csc (4 x)) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1 - v ^{2}}{v ^{4}} d v

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

Expand

\frac{1 - v ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{1}{v ^{2}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int \frac{1}{v ^{4}} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

= \frac{1}{4} (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

Substitute back

= \frac{1}{4} (- \frac{1}{3 sin ^{3} ( 4 x )} - (- \frac{1}{sin ( 4 x )}))

Simplify

\frac{1}{4} (- \frac{1}{3 sin ^{3} ( 4 x )} - (- \frac{1}{sin ( 4 x )})) : \frac{1}{4} (- \frac{1}{3} csc^{3} (4 x) + csc (4 x))

= \frac{1}{4} (- \frac{1}{3} csc^{3} (4 x) + csc (4 x))

Add a constant to the solution

= \frac{1}{4} (- \frac{1}{3} csc^{3} (4 x) + csc (4 x)) + C

\frac{d}{d x} ((x - \frac{1}{x})^{3})

\int \frac{7 x ^{3} + 7 x}{x - 1} d x

y^{^{'}} + 5 y = t + e^{- 2 t}

y = 7 x - x^{2}, y = 2 x

\int_{1}^{4} (3 x + 5) d x

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