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

The integral of (csc(4x)+cot(4x))^2 is 1/2 (-cot(2x)-arctan(tan(2x)))+C

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

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

\frac{1}{2} (- cot (2 x) - arctan (tan (2 x))) + C

Solution steps

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

Apply u-substitution

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

Apply u-substitution

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{4} \cdot 2 (- \frac{1}{v} - arctan (v))

Substitute back

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

Simplify

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

= \frac{1}{2} (- cot (2 x) - arctan (tan (2 x)))

Add a constant to the solution

= \frac{1}{2} (- cot (2 x) - arctan (tan (2 x))) + C

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