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

The integral of cot^4(3x)csc^2(3x) is -1/15 cot^5(3x)+C

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

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

- \frac{1}{15} cot^{5} (3 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - v^{4} d v

= \frac{1}{3} \cdot \int - v^{4} 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^{4} d v)

Apply the Power Rule

: \frac{v ^{5}}{5}

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

Substitute back

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

Simplify

\frac{1}{3} (- \frac{cot ^{5} ( 3 x )}{5}) : - \frac{1}{15} cot^{5} (3 x)

= - \frac{1}{15} cot^{5} (3 x)

Add a constant to the solution

= - \frac{1}{15} cot^{5} (3 x) + C

\int [cot (x) - csc (x)]^{2} d x

\int \frac{x + 3}{x ^{2} ( x ^{2} + 9 )} d x

\int \frac{( 2 x ^{2} + 3 )}{x ^{3} - 2 x ^{2} + x} d x

\int \frac{1}{1 - cot ( x )} d x

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

What is the integral of cot^4(3x)csc^2(3x) ?
The integral of cot^4(3x)csc^2(3x) is -1/15 cot^5(3x)+C