What is the integral of (tan(3x)-cot(3x))^3 ?

The integral of (tan(3x)-cot(3x))^3 is -4/3 (-1/2 cot^2(6x)-ln|sin(6x)|)+C

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integral of (tan(3x)-cot(3x))^3

\int (tan (3 x) - cot (3 x))^{3} d x

- \frac{4}{3} (- \frac{1}{2} cot^{2} (6 x) - ln ∣ sin (6 x) ∣) + C

Solution steps

\int (tan (3 x) - cot (3 x))^{3} d x

Apply u-substitution

: \int (- cot (u) + tan (u))^{3} \frac{1}{3} d u

= \int (- cot (u) + tan (u))^{3} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (- cot (u) + tan (u))^{3} d u

Simplify

(- cot (u) + tan (u))^{3} : - 8 cot^{3} (2 u)

= \frac{1}{3} \cdot \int - 8 cot^{3} (2 u) d u

Take the constant out:

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

= \frac{1}{3} (- 8 \cdot \int cot^{3} (2 u) d u)

Apply u-substitution

: \int cot^{3} (v) \frac{1}{2} d v

= \frac{1}{3} (- 8 \cdot \int cot^{3} (v) \frac{1}{2} d v)

Take the constant out:

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

= \frac{1}{3} (- 8 \cdot \frac{1}{2} \cdot \int cot^{3} (v) d v)

Apply Integral Reduction

= \frac{1}{3} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - \int cot (v) d v))

Rewrite using trig identities

= \frac{1}{3} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - \int \frac{cos ( v )}{sin ( v )} d v))

Apply u-substitution

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

= \frac{1}{3} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - \int \frac{1}{w} d w))

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{3} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - ln ∣ w ∣))

Substitute back

= \frac{1}{3} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( 2 \cdot 3 x )}{2} - ln ∣ sin (2 \cdot 3 x) ∣))

Simplify

\frac{1}{3} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( 2 \cdot 3 x )}{2} - ln ∣ sin (2 \cdot 3 x) ∣)) : - \frac{4}{3} (- \frac{1}{2} cot^{2} (6 x) - ln ∣ sin (6 x) ∣)

= - \frac{4}{3} (- \frac{1}{2} cot^{2} (6 x) - ln ∣ sin (6 x) ∣)

Add a constant to the solution

= - \frac{4}{3} (- \frac{1}{2} cot^{2} (6 x) - ln ∣ sin (6 x) ∣) + C

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What is the integral of (tan(3x)-cot(3x))^3 ?
The integral of (tan(3x)-cot(3x))^3 is -4/3 (-1/2 cot^2(6x)-ln|sin(6x)|)+C