What is the integral of (csc^4(2x))/(cot^2(2x)) ?

The integral of (csc^4(2x))/(cot^2(2x)) is -cot(4x)+C

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integral of (csc^4(2x))/(cot^2(2x))

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

- cot (4 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{csc ^{4} ( u )}{2 cot ^{2} ( u )} d u

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{csc ^{4} ( u )}{cot ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{8}{1 - cos ( 4 u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot 8 \cdot \int \frac{1}{1 - cos ( 4 u )} d u

Apply u-substitution

: \int \frac{1}{4 ( 1 - cos ( v ) )} d v

= \frac{1}{2} \cdot 8 \cdot \int \frac{1}{4 ( 1 - cos ( v ) )} d v

Take the constant out:

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

= \frac{1}{2} \cdot 8 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( v )} d v

Apply u-substitution

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

= \frac{1}{2} \cdot 8 \cdot \frac{1}{4} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= \frac{1}{2} \cdot 8 \cdot \frac{1}{4} (- \frac{1}{w})

Substitute back

= \frac{1}{2} \cdot 8 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 \cdot 2 x}{2} )})

Simplify

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

= - cot (4 x)

Add a constant to the solution

= - cot (4 x) + C

x \frac{d y}{d x} = y + (x^{2} - 1)^{2}

x y^{^{'}} + y = - 9 x y^{2}, y (1) = - 2

x \to x + 2 lim (2 x^{2} + 3 x - 2)

(\frac{sin ( x )}{x})^{^{'}}

f (x) = ln (3 + 4 x)

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