What is the integral of csc^3(4x) ?

The integral of csc^3(4x) is 1/32 (-cot^2(2x)+4ln|tan(2x)|+tan^2(2x))+C

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integral of csc^3(4x)

\int csc^{3} (4 x) d x

\frac{1}{32} (- cot^{2} (2 x) + 4 ln ∣ tan (2 x) ∣ + tan^{2} (2 x)) + C

Solution steps

\int csc^{3} (4 x) d x

Apply u-substitution

= \int csc^{3} (u) \frac{1}{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 csc^{3} (u) d u

Apply u-substitution

= \frac{1}{4} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{4 v ^{3}} d v

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{4} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} d v

Expand

\frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} : \frac{1}{v ^{3}} + \frac{2}{v} + v

= \frac{1}{4} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{3}} + \frac{2}{v} + v 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} \cdot \frac{1}{4} (\int \frac{1}{v ^{3}} d v + \int \frac{2}{v} d v + \int v d v)

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

\int \frac{2}{v} d v = 2 ln ∣ v ∣

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

= \frac{1}{4} \cdot \frac{1}{4} (- \frac{1}{2 v ^{2}} + 2 ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= \frac{1}{4} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{4 x}{2} )} + 2 ln tan (\frac{4 x}{2}) + \frac{tan ^{2} ( \frac{4 x}{2} )}{2})

Simplify

\frac{1}{4} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{4 x}{2} )} + 2 ln tan (\frac{4 x}{2}) + \frac{tan ^{2} ( \frac{4 x}{2} )}{2}) : \frac{1}{32} (- cot^{2} (2 x) + 4 ln ∣ tan (2 x) ∣ + tan^{2} (2 x))

= \frac{1}{32} (- cot^{2} (2 x) + 4 ln ∣ tan (2 x) ∣ + tan^{2} (2 x))

Add a constant to the solution

= \frac{1}{32} (- cot^{2} (2 x) + 4 ln ∣ tan (2 x) ∣ + tan^{2} (2 x)) + C

What is the integral of csc^3(4x) ?
The integral of csc^3(4x) is 1/32 (-cot^2(2x)+4ln|tan(2x)|+tan^2(2x))+C