What is the integral of cot^2(3x)csc^4(3x) ?

The integral of cot^2(3x)csc^4(3x) is -1/3 (1/5 cot^5(3x)+1/3 cot^3(3x))+C

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integral of cot^2(3x)csc^4(3x)

\int cot^{2} (3 x) csc^{4} (3 x) d x

- \frac{1}{3} (\frac{1}{5} cot^{5} (3 x) + \frac{1}{3} cot^{3} (3 x)) + C

Solution steps

\int cot^{2} (3 x) csc^{4} (3 x) d x

Apply u-substitution

: \int cot^{2} (u) csc^{4} (u) \frac{1}{3} d u

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Expand

v^{2} (v^{2} + 1) : v^{4} + v^{2}

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

\int v^{4} d v = \frac{v ^{5}}{5}

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

= \frac{1}{3} (- (\frac{v ^{5}}{5} + \frac{v ^{3}}{3}))

Substitute back

= \frac{1}{3} (- (\frac{cot ^{5} ( 3 x )}{5} + \frac{cot ^{3} ( 3 x )}{3}))

Simplify

\frac{1}{3} (- (\frac{cot ^{5} ( 3 x )}{5} + \frac{cot ^{3} ( 3 x )}{3})) : - \frac{1}{3} (\frac{1}{5} cot^{5} (3 x) + \frac{1}{3} cot^{3} (3 x))

= - \frac{1}{3} (\frac{1}{5} cot^{5} (3 x) + \frac{1}{3} cot^{3} (3 x))

Add a constant to the solution

= - \frac{1}{3} (\frac{1}{5} cot^{5} (3 x) + \frac{1}{3} cot^{3} (3 x)) + C

x \to 0 lim (x ln (sin (x)))

\int 4 e^{3 x} cos (5 x) d x

\int_{1}^{16} 4 x d x

ln (x + 2)

\int_{0}^{1} (5 x)^{2} - (5 x^{4})^{2} d x

What is the integral of cot^2(3x)csc^4(3x) ?
The integral of cot^2(3x)csc^4(3x) is -1/3 (1/5 cot^5(3x)+1/3 cot^3(3x))+C