What is the integral of cot^5(2bx) ?

The integral of cot^5(2bx) is -1/(2b)(-1/2 cot^2(2bx)+1/4 cot^4(2bx)+1/2 ln(csc^2(2bx)))+C

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integral of cot^5(2bx)

\int cot^{5} (2 b x) d x

- \frac{1}{2 b} (- \frac{1}{2} cot^{2} (2 b x) + \frac{1}{4} cot^{4} (2 b x) + \frac{1}{2} ln (csc^{2} (2 b x))) + C

Solution steps

\int cot^{5} (2 b x) d x

Apply u-substitution

= \int \frac{cot ^{5} ( u )}{2 b} d u

Take the constant out:

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

= \frac{1}{2 b} \cdot \int cot^{5} (u) d u

Apply u-substitution

= \frac{1}{2 b} \cdot \int - \frac{v ^{5}}{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}{2 b} (- \int \frac{v ^{5}}{1 + v ^{2}} d v)

Long division

\frac{v ^{5}}{1 + v ^{2}} : v^{3} - v + \frac{v}{v ^{2} + 1}

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

\int v^{3} d v = \frac{v ^{4}}{4}

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

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

= \frac{1}{2 b} (- (\frac{v ^{4}}{4} - \frac{v ^{2}}{2} + \frac{1}{2} ln v^{2} + 1))

Substitute back

= \frac{1}{2 b} (- (\frac{cot ^{4} ( 2 b x )}{4} - \frac{cot ^{2} ( 2 b x )}{2} + \frac{1}{2} ln cot^{2} (2 b x) + 1))

Simplify

\frac{1}{2 b} (- (\frac{cot ^{4} ( 2 b x )}{4} - \frac{cot ^{2} ( 2 b x )}{2} + \frac{1}{2} ln cot^{2} (2 b x) + 1)) : - \frac{1}{2 b} (- \frac{1}{2} cot^{2} (2 b x) + \frac{1}{4} cot^{4} (2 b x) + \frac{1}{2} ln (csc^{2} (2 b x)))

= - \frac{1}{2 b} (- \frac{1}{2} cot^{2} (2 b x) + \frac{1}{4} cot^{4} (2 b x) + \frac{1}{2} ln (csc^{2} (2 b x)))

Add a constant to the solution

= - \frac{1}{2 b} (- \frac{1}{2} cot^{2} (2 b x) + \frac{1}{4} cot^{4} (2 b x) + \frac{1}{2} ln (csc^{2} (2 b x))) + C

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